A Potpourri of a Diverse Variety of Algebra Problems(including a Crux Mathematicorum Olympiad...

A Potpourri of a Diverse Variety of Algebra Problems (including a Crux

Mathematicorum Olympiad Corner Problem): Fourteen problems in total

Konstatine Zelator

Department of Mathematics

317 Bishop Hall

SUNY Buffalo State College

1300 Elmwood Ave

Buffalo, NY 14222 USA

[email protected]

Also:

K. Zelator

P.O. Box 4280

Pittsburgh, PA 15203 USA

konstantine [email protected]

1. Introduction

As the title of this article points out, there is indeed a diverse variety of algebra problems in this

paper. This is a compilation of fourteen algebra problems from four different sources.

Problem 14, is an Olympiad Corner Problem, OC 91, published in the September 2012 issue of

the journal, Crux Mathematicorum with Mathematical Mayhem(see reference [1]). Problem 11 can

be found in the linear algebra text, Elementary Linear Algebra Applications Version, by Howard

Anton (see reference [2]).

Problem 5 is of this author’s own making. The rest of the problems were originally found in

an obscure classical algebra book, published circa 1971 in Athens, Greece. However, some of these

problems have been modified by this author (see reference [3]).

In Section 2, we list all fourteen problems.

In Section 3, the reader will find detailed solutions to these problems.

1

2. Listing of Problems

Problem 1: Find the positive integer solutions to the 4-variable system:

»

—

—

–

x`yiz`wi “ 3, z ` 3w “ 20

|px` yiq ` pz ` wiq| ă |x` yi| ´ |z ` wi| `?

10zw

fi

ffi

ffi

fl

; where i is the imaginary unit i “?´1

Problem 2: Find the real values of k such that the quadratic equation x2 ` p2k ` 1qx ` k ` 4 “ 0

has two distinct real roots r1, r2 such that r1 ą r2 and |r1| ´ 2|r2| “ k ` 1.

Problem 3: Let a, b, c be real numbers. consider the 3-variable linear system:

pa´ cqz ´ pb´ aqy “ 3a

pb´ aqx´ pc´ bqz “ 3b

pc´ bqy ` pa´ cqx “ 3c

(i) Show that if the three real numbers a, b, c are distinct, then the above system has a unique

solution. Find that solution.

(ii) Suppose that a, b, c are integers such that a and c are equal, but distinct from b (a “ b ‰ c).

Show that there is a specific condition that the integers a and b must satisfy in order that the above

has an integer solution. Show that if a and b satisfy that condition, the system has infinitely many

integer solutions. Describe the solution set, in integers x, y, and z. Otherwise, if a and b do not

satisfy the said condition, the above system has no integer solution.

2

Problem 4: Let d and k be given positive integers. Consider a triangle that satisfies the following

two conditions.

1. The three side lengths of the triangle are consecutive terms of an arithmetic progression with

common difference d.

2. Let AE “ k; where A is the area of the triangle, and E being the area of the rectangle who two

pairs of side lengths are the two smallest side lengths (among the three) of the triangle.

For which values of k does this problem have no solution?

For which values of k does this problem have a unique solution? Find the three side-lengths.

For which values of k does this problem have two solutions? Find the three side-lengths.

Problem 5: Let a, b be real numbers. Consider the two polynomial functions:

fpxq “ x3 ` 2pa´ bqx2 ` ax` 15´ 2pa´ bq; and

gpxq “ x2 ` pa´ bqx` b´ 2

If the polynomial gpxq divides fpxq, find the three complex roots of fpxq (which will have either

three real roots, or one real root and two non-real conjugate complex roots).

Problem 6: Find the integer values of the constant k, for which the equation x3´pk` 1qx` 2k “ 0

has a rational root.

Problem 7: Consider the one-parameter family of polynomial functions, fpxq “ x4´pp` 3qx2` p2.

Find the values of the real parameter p, such that fpxq has four real roots.

3

Problem 8: Solve the following equation:

Let a ą 0, a ‰ 1; logarloga xs “ loga2 rloga2 xs

Problem 9: Determine all the real solutions of the following 3-variable system:

”

1x `

1y `

1z “ 1, xyz “ 2, z ă y ă 0 ă x

ı

Problem 10: Let p and q be rational numbers. Consider the quadratic trinomials t1pxq “ x2´px`q

and t2pxq “ x2 ` qx´ p. Suppose that these two trinomials have only one root in common. Show

that the trinomial whose roots are the other two roots (of t1pxq and t2pxq) has rational roots.

Problem 11: (see reference [2]) For which values of the constant c will the following linear system

of three equations, in three unknowns x, y, z; have no solutions? Exactly one solution? Infinitely

many solutions?

»

—

—

—

—

—

—

–

x` 2y ´ 3z “ 4

3x´ y ` 5z “ 2

4x` y ` pc2 ´ 14qz “ c` 2

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

Problem 12: Determine all rectangles with integer side lengths and with the property that their

areas are four times the sum of half their perimeters and one of the their diagonals.

4

Problem 13: Find all the positive integer solutions of the 4-variable system of equations:

»

—

—

—

—

—

—

–

x` y “ 100

xy “ z2

1x `

1y “

2w

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

Problem 14: (see reference [1]) Prove that no integer consisting of one 2, one 1, and the remaining

digits all 0 can be written either as the sum of two perfect cubes or the sum of two perfect squares.

5

3. Solutions to Problems

Problem 1 We have:

|px` yiq ` pz ` wiq| “ |px` zq ` py ` wqi| “a

px` zq2 ` py ` wq2;

|x` yi| “a

x2 ` y2, |z ` wi| “?z2 ` w2.

So that the given system is equivalent to,

tx` yi “ 3z ` 3wi; z ` 3w “ 20;a

px` zq2 ` py ` wq2 ăa

x2 ` y2 ´?z2 ` w2 `

?10zwu

ÐÑ

tx “ 3z, y “ 3w, z “ 20´ 3w; and 4?z2 ` w2 ă 3

?z2 ` w2 ´

?z2 ` w2 `

?10zwu

ÐÑ

tx “ 3z, y “ 3w, z “ 20´ 3w; and 2?z2 ` w2 ă

?10zwu

ÐÑ

tx “ 3z, y “ 3w, z “ 20´ 3w; and 2pz2 ` w2q ă 5zwu (1)

Since z and w are positive integers; it follows from equation (1) that,

w “ 1, 2, 3, 4, 5, 6; and correspondingly, z “ 17, 14, 11, 8, 5, or 2.

Of these six possible solutions, only the pair pz, wq “ p5, 5q, satisfies the inequality in (1). Therefore,

this problem has a unique solution:

(x,y,z,w)=(15,15,5,5)

6

Problem 2 Quadratic equation (with the roots satisfying |r1| ´ 2|r2| “ k ` 1q

x2 ` p2k ` 1qx` k ` 4 “ 0 (1)

The condition that equation (1) has two distinct real roots means that its discriminant must be

positive:

p2k ` 1q2 ´ 4pk ` 4q ą 0

ÐÑ

4k2 ` 4k ` 1´ 4k ´ 16 ą 0

ÐÑ

4k2 ą 15

ÐÑ

2|k| ą?

15

ÐÑ

k ą?152 or k ă ´

?152 (2)

Using the quadratic formula, we see that the two roots of equation (1) are the numbers;

´p2k`1q`?4k2´15

2 and ´p2k`1q´?4k2´15

2 ;

and since?

4k2 ´ 15 ą 0; the first root is the greater of the two:

r1 “´p2k`1q`

?4k2´15

2 ą r2 “´rp2k`1q`

?4k2´15s

2 (3)

Case 1: Suppose that k ą?152 in (2). Then, obviously, 2k ` 1 ą 0 and k ` 4 ą 0 Therefore,

the sum of the two roots is r1 ` r2 “ ´p2k ` 1q ă 0 and the product r1r2 “ k ` 4 ą 0. Therefore,

one of the r1, r2; must be positive, the other negative. This implies by (3) that:

r1 ą 0 and r2 ă 0.

7

Thus, r1 ą 0

ÐÑ?

4k2 ´ 15 ą 2k ` 1 ą 0;

ÐÑ 4k2 ´ 15 ą p2k ` 1q2;

ÐÑ 4k2 ´ 15 ą 4k2 ` 4k ` 1; k ă ´4.

This contradicts the assumption that k ą?152 , ruling out Case 1.

Case 2: Assume the second possibility in (2), namely that k ă ´?152 .

This implies that 2k`?

15 ă 0 ÐÑ 2k`1 ă ´?

15`1 ă 0. And so, in this case we have 2k`1 ă 0;

or equivalently,

´p2k ` 1q ą 0 (4)

On the other hand,

k ` 4 ă ´?152 ` 4;

ÐÑ

k ` 4 ă 8´?15

2

And since, 8´?15

2 ą 0; the last inequality shows that k ` 4 can be positive, zero, or negative (of

course, without considering additional information). So we have three sub-cases:

Sub-Case 2a - Suppose that:

0 ă k ` 4 (5)

Since r1`r2 “ ´p2k`1q ą 0 (by (4)) and 0 ă r1r2 “ k`4 (by (5)), it follows that in this sub-case,

both roots r1 and r2 must be positive. But by (3), just r2 ą 0; implies that r1 ą 0. We have r2 ą 0,

which by (3) is equivalent to:

´p2k ` 1q ą?

4k2 ´ 15 ą 0

ÐÑ

8

r´p2k ` 1qs2 ą 4k2 ´ 15

ÐÑ

4k2 ` 4k ` 1 ą 4k2 ´ 15

ÐÑ

k ą ´4; k ` 4 ą 0, confirming condition (5) .

Next, apply the condition |r1|´2|r2| “ k`1. Since r1 ą 0 and r2 ą 0, we obtain |r1| “ r1, |r2| “ r2.

And so,

r1 ´ 2r2 “ k ` 1 (6)

Applying (3) and (6) gives:

´p2k`1q`?4k2´15

2 ´ 2”

´p2k`1q´?4k2´15

2

ı

“ k ` 1 (7)

Equation (7), when simplified, yields;

3?

4k2 ´ 15 “ 1;

4k2 ´ 15 “ 19 ;

4k2 “ 1369 ;

k2 “ 349 ;

|k| “?343 (8)

But, k ă 0, since k ă ´?152 by Case 2.

Thus, by (8), k “ ´?343 . It is also true that:

´?343 ă ´

?152

ÐÑ

?343 ą

?152

ÐÑ

2?

34 ą 3?

15

9

ÐÑ

22 ˚ p34q ą 32 ˚ p15q

ÐÑ

136 ą 135

Also,

0 ă k ` 4,´4 ă k,´4 ă ´?343

ÐÑ

12 ą?

34, p12q2 ą 34

Therefore, we obtain the solution; k “ ´?343

´

´4 ă ´?343 ă ´

?152

¯

Sub-Case 2b - Assume k ` 4 “ 0 and k “ ´4. A calculation in (3) shows that r1 “ 4 and

r2 “ 3. But then, r1 ´ 2r2 “ 4´ 6 “ ´2 ‰ ´3 “ k ` 1. Therefore, this sub-case is ruled out.

Sub-Case 2c - Assume that k ` 4 ă 0; k ă ´4. Then, 2k ` 1 ă p´8 ` 1 “ ´7q ă 0 and k ` 4 ă 0

so that r1 ` r2 “ ´p2k ` 1q ą 7 ą 0 and r1r2 “ k ` 4 ă 0. This shows, by (3), that we must have

r1 ą 0 and r2 ă 0.

Apply the condition |r1| ´ 2|r2| “ k ` 1: We have |r1| “ r1 and |r2| “ ´r2.

And so,

r1 ` 2r2 “ k ` 1;

´p2k`1q`?4k2´15

2 ` 22

“

´p2k ` 1q ´?

4k2 ´ 15‰

“ k ` 1;

´3p2k ` 1q ´?

4k2 ´ 15 “ 2pk ` 1q;

´?

4k2 ´ 15 “ 8k ` 5 (note that 8k ` 5 ă ´27 since k ă ´4) (8)

10

Keeping in mind that k ă ´4, squaring both sides of equation (8) yields:

4k2 ´ 15 “ 64k2 ` 80k ` 25;

60k2 ` 80k ` 40 “ 0;

ÐÑ

3k2 ` 4k ` 2 “ 0,

which has two conjugate complex solutions (since the discriminant of this last quadratic equation

in k is 16´ 24 “ ´8 ă 0).

3k2 ` 4k ` 2 “ 3`

k ` 23

˘2` 2

3 ě23 ą 0 .

Thus, Sub-Case 2c is ruled out as well.

Conclusion: Problem 2 has a unique solution:

k “ ´?343

11

Problem 3 We have:

pa´ bqy ` pa´ cqz “ 3a (1)

pb´ aqx ` pb´ cqz “ 3b (2)

pa´ cqx` pc´ bqy “ 3c (3)

(i) We apply Cramer’s Rule for linear systems of n equations with n unknowns; in our case n “ 3.

Let D be the determinant of the matrix of the coefficients.

Then D “

»

—

—

—

—

—

—

–

0 (a-b) (a-c)

(b-a) 0 (b-c)

(a-c) (c-b) 0

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

D “ ´pa´ bq

»

—

—

–

pb´ aq pb´ cq

pa´ cq 0

fi

ffi

ffi

fl

` pa´ cq

»

—

—

–

pb´ aq 0

pa´ cq pc´ bq

fi

ffi

ffi

fl

;

D “ pb´ aq r0´ pa´ cqpb´ cqs ` pa´ cq rpb´ aqpc´ bq ´ 0s;

D “ ´pb´ aqpa´ cqpb´ cq ` pa´ cqpb´ aqpc´ bq;

D “ pa´ bqpa´ cqpb´ cq ` pa´ cqp´1qpa´ bqp´1qpb´ cq;

D “ 2pa´ bqpa´ cqpb´ cq ‰ 0, since a, b, c are distinct.

Therefore, the system has a unique solution.

12

Next, we compute the determinant, Dx. We have:

Dx “

»

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—

—

—

—

–

3a pa´ bq pa´ cq

3b 0 pb´ cq

3c pc´ bq 0

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

;

Dx “ 3a

»

—

—

–

0 pb´ cq

pc´ bq 0

fi

ffi

ffi

fl

´ pa´ bq

»

—

—

–

3b pb´ cq

3c 0

fi

ffi

ffi

fl

` pa´ cq

»

—

—

–

3b 0

3c pc´ bq

fi

ffi

ffi

fl

;

Dx “ 3a p0´ pc´ bqpb´ cqq ´ pa´ bq p0´ 3cpb´ cqq ` pa´ cq p3bpc´ bq ´ 0q;

Dx “ 3apb´ cq2 ` 3cpa´ bqpb´ cq ` 3bpa´ cqpc´ bq;

Dx “ 3pb´ cq rapb´ cq ` cpa´ bq ´ bpa´ cqs;

Dx “ 3pb´ cq rab´ ac` ca´ cb´ ba` cbs “ 0, pab´ ac` ca´ cb´ ba` cbq “ 0;

Thus, x “ DxD “ 0

2pa´bqpaćqpbćq “ 0

Instead of continuing with Cramer’s Rule, in order to find y and z; y “Dy

D , z “ DzD . We simply go

back to the system of equations:

From equation (2), with x “ 0; we obtain z “ 3bbć , and from equation (3), y “ 3c

c´b .

Let us check equation (1):

pa´ bqp 3cc´bq ` pa´ cqp 3b

bćq “3cpa´bq´paćq3b

c´b “ 3ca´3cb´3ab`3cbc´b “

3apc´bqc´b “ 3a

The unique solution to the system is px, y, zq “´

0, 3cc´b ,

3bbć

¯

.

13

(ii). Suppose that a “ c ‰ b; where a, b P Z. Then the given system takes the form:

pa´ bqy “ 3a (4)

pb´ aqpx` zq “ 3b (5)

Since a ‰ b; |a ´ b| is a positive integer. Let d be the greatest common divisor (gcd) of a and

b; 1 ď d “gcdpa, bq. Then,

a “ d ˚ a1 and b “ d ˚ b1; where a1 and b1 are relatively prime integers; gcdpa1, b1q “ 1 (6)

From (4), (5), and (6); we obtain:

pa1 ´ b1q ˚ y “ 3a1 (7)

pb1 ´ a1qpx` zq “ 3b1 (8)

So, if x, y, x are integers; then (7) and (8) imply that the positive integer |a1 ´ b1| is a common

divisor of 3a1 and 3b1:

p|a1 ´ b1|q � 3a1 and p|a1 ´ b1|q � 3b1 (9)

Moreover, (9) implies that |a1 ´ b1| must be a divisor of the greatest common divisor of 3a1 and

3b1:

p|a1 ´ b1|q � gcdp3a1, 3b1q “ 3gcdpa1, b1q “ 3 ˚ 1 “ 3 by (6).

14

Therefore, |a1 ´ b1| “ 1 or 3. This is the precise condition:

Let d “gcdpa, bq, a1 “ad , b1 “

bd . If (and only if) the integers a1 and b1 satisfy the condition |a1 ´ b1| “ 1 or 3,

the given system will have integer solutions. Otherwise, it will have no integer solution.

Finding The Integer Solutions

Going back to the system of (7) and (8); we easily see that when the above condition is satisfied.

The system will have infinitely many solutions; which can be parametrically expressed in terms of

parameter ptq as follows:

Let t P Z, (t “ 0,˘1,˘2, ....).

1. If a1 ´ b1 “ 1, then the solution set S can be described by x “ ´3b1 ´ t, y “ 3p1` b1q, z “ t.

2. If a1 ´ b1 “ ´1, then the solution set S can be described by x “ 3b1 ´ t, y “ 3p1´ b1q, z “ t.

3. If a1 ´ b1 “ 3, then the solution set S can be described by x “ ´b1 ´ t, y “ b1 ` 3, z “ t.

4. If a1 ´ b1 “ ´3, then the solution set S can be described by x “ b1 ´ t, y “ 3´ b1, z “ t.

15

Problem 4 We make use of the well-known Heron’s Formula for the area of a triangle. If a, b, c are

side lengths of a triangle, then the area A is given by,

A “a

s ˚ ps´ aqps´ bqps´ cq;

where s is the semi-perimeter; s “ a`b`c2 .

Using this, we see that the area formula can be written in the form:

4A “a

pa` b` cqpb` c´ aqpa` c´ bqpa` b´ cq;

or, equivalently,

16A2 “ pa` b` cqpb` c´ aqpa` c´ bqpa` b´ cq (1)

Now consider the information of the problem. Let x ą 0 be the smallest of the three side lengths of

the triangle. Then, since the three side lengths are consecutive terms of an arithmetic progression

with common difference d; the other two side lengths are x` d and x` 2d, pd ą 0q.

For a triangle to be formed, the three triangle inequalities must be satisfied (necessary and sufficient

condition):¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

x` px` dq ą x` 2d;

x` px` 2dq ą px` dq;

px` dq ` px` 2dq ą x;

with d ą 0

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

Ø

¨

˚

˚

˚

˚

˚

˚

˝

x ą d ą 0;

x` d ą 0;

x` 3d ą 0

˛

‹

‹

‹

‹

‹

‹

‚

Ø px ą d ą 0q (2)

Next, take a “ x, b “ x` d, c “ x` 2d; and apply (1):

16A2 “ 3px` dqpx` 3dqpx` dqpx´ dq;

16A2 “ 3px` dq2px` 3dqpx´ dq (3)

The rectangle with two pairs of side lengths x and x` d has area E “ xpx` dq. According to the

problem’s hypothesis, we have:

AE “ k; A “ kxpx` dq (4)

16

Combining (3) and (4) gives:

16k2x2px` dq2 “ 3px` dq2px` 3dqpx´ dq Ø px` dq2“

16k2x2 ´ 3px` 3dqpx´ dq‰

“ 0.

However, x ą 0, d ą 0; which implies that px` dq2 ą 0; px` dq2 ‰ 0. And so, the last equation is

equivalent to the quadratic equation, 16k2x2´ 3px` 3dqpx´ dq “ 0; which reduces to the equation,

p16k2 ´ 3qx2 ´ 6dx` 9d2 “ 0 (5)

Case 1 If 16k2 ´ 3 “ 0; 0 ă k “?34 , equation (5) becomes the linear equation ´6dx ` 9d2 “ 0,

which has the solution x “ 3d2 .

Note that 3d2 ą d ą 0; so condition (2) is satisfied. The other two side lengths are:

x` d “ 3d2 ` d “ 5d

2 .

x` 2d “ 3d2 ` 2d “ 7d

2 .

In this case, k “?34 ; the solution to the problem is the triangle with side lengths 3d

2 , 5d2 , 7d

2 . This

triangle is similar to the obtuse triangle with side lengths 3, 5, and 7.

Case 2: 16k2 ´ 3 ‰ 0; k ‰?34

In this case, (5) is a genuine quadratic equation. We compute the discriminant D.

D “ p´6dq2 ´ 4p16k2 ´ 3qp9d2q “ 36d2p1´ 16k2 ` 3q;

D “ 36d2p4´ 16k2q “ 36d2 ˚ 42`

14 ´ k2

˘

;

D “`

12d2˘ `

12 ` k

˘ `

12 ´ k

˘

(6)

If k ą 12 , then by (6), we have D ă 0; and equation (5) has no real number solutions; it has two

conjugate complex roots. Therefore, the problem has no solution in this case.

We have 0 ă k ă 12 and k ‰

?34 ă 1

2 . So if, 0 ă k ă?34 ă 1

2 , then the root r1 is negative, but r2 is

positive. Indeed, 0 ă k ă?34 implies 16k2 ´ 3 ă 0. This shows, by the formula below, that r1 ă 0

(since d ą 0).

17

But then, 1´ 2?

1´ 4k2 ă 0 Ø 1 ă 2?

1´ 4k2 Ø 1 ă 4p1´ 4k2q Ø 16k2 ´ 3 ă 0.

Thus, both the numerator and denominator in the formula for r2 below are negative; so r2 ą 0

Thus, the value of the side length in this case is,

x “ r2 “3dp1´2

?1´4k2q

16k2´3“

3dp2?1´4k2´1q

3´16k2.

Is x ą d, as required by (2)? We have:

3dp2?1´4k2´1q

3´16k2ą d

ÐÑ

3dp2?

1´ 4k2 ´ 1q ą dp3´ 16k2q; and since d ą 0

ÐÑ

6?

1´ 4k2 ´ 3 ą 3´ 16k2

ÐÑ

6?

1´ 4k2 ą 6´ 16k2

ÐÑ

3?

1´ 4k2 ą 3´ 8k2 ą 0 (since, k2 ă 316 ă

38)

squaring both sides

ÐÑ

9p1´ 4k2q ą 9´ 48k2 ` 64k4

ÐÑ

0 ą 64k4 ´ 12k2;

ÐÑ

0 ą 64k2pk2 ´ 316q, which is true

On the other hand, if k “ 12 , then D “ 0; and so (5) has a real root in this sub-case. We

18

have 16k2 ´ 3 “ 16`

12

˘2´ 3 “ 4 ´ 3 “ 1. So the double real root is x “ 6d

2 “ 3d; note that

x “ 3d ą d ą 0, so condition (2) is satisfied. The other two side lengths are 4d and 5d.

In this case, k “ 12 ; the solution to the problem is the triangle with side lengths

3d, 4d, 5d; similar to the ninety degree triangle with side lengths 3, 4, 5.

Note that?34 ă 1

2 , since?

3 ă 2.

Back to Equation (5), with 0 ă k ă?34 , Equation (5) has two distinct real roots, then by the

quadratic formula:

r1 “6d`

?D

2p16k2´3qand r2 “

6d´?D

2p16k2´3q(7)

Using D “ 36d2p4´ 16k2q “ p12dq2 ˚ p1´ 4k2q from (6);

Formulas (7) yield (when simplified):

r1 “3dp1`2

?1´4k2q

16k2´3, r2 “

3dp1´2?1´4k2q

16k2´3(8)

4k2p16k2 ´ 3q ă 0, which is true since 0 ă k ă?34 . So yes, the condition (2) is satisfied. So

in this case, we obtain the triangle r2, r2 ` d, r2 ` 2d; where r2 “3dp2

?1´4k2´1q

3´16k2.

Finally, suppose that k satisfies?34 ă k ă 1

2 . Then both 16k2 ´ 3 and 1 ´ 2?

1´ 4k2 are pos-

itive. Thus, both r1 and r2 are positive in this case, as (8) clearly shows. So this problem has

two solutions in this sub-case (of Case 2): r1, r1 ` d, r1 ` 2d and r2, r2 ` d, r2 ` 2d (in this case,

0 ă r2 ă r2). One verifies again that r2 ą d (similar as before, in the case 0 ă k ă?34 ; since both

numerator and denominator are now positive), so 0 ă d ă r2 ă r1.

19

Summary of Results

1. If k ą 12 , the problem has no solution.

2. If k “ 12 , the problem has the unique solution 3d, 4d, 5d (the three side lengths).

3. If k “?34 the problem has the unique solution 3d

2 ,5d2 ,

7d2 .

4. If 0 ă k ă?34 , the problem has the unique solution r2, r2`d, r2`2d; where r2 “

3dp2?1´4k2´1q

3´16k2.

5. If?34 ă k ă 1

2 , the problem has two solutions: r2, r2` d, r2` 2d and r1, r1` d, r1` 2d, where:

r2 “3dp1´2

?1´4k2q

16k2´3and r1 “

3dp1`2?1´4k2q

16k2´3

20

Problem 5:

We have fpxq “ x3 ` 2pa´ bqx2 ` ax` 15´ 2pa´ bq and gpxq “ x2 ` pa´ bqx` b´ 2.

The quotient qpxq of the division of fpxq and gpxq must be a degree one polynomial; qpxq “ kx`m;

k,m P R, k ‰ 0. Since, by hypothesis, gpxq divides fpxq; the remainder of the division is zero. Thus,

fpxq “ qpxqgpxq;

x3 ` 2pa´ bqx2 ` ax` 15´ 2pa´ bq “ pkx`mqpx2 ` pa´ bqx` b´ 2q

Which, after some algebra, reduces to:

»

—

—

–

x3 ` 2pa´ bqx2 ` ax` 15´ 2pa´ bq “

kx3 ` pm` kpa´ bqqx2 ` rmpa´ bq ` kpb´ 2qsx`mpb´ 2q

fi

ffi

ffi

fl

(1)

Since the two polynomial functions on either side of (1) are equal or identical, according to well-

known result in algebra, their corresponding coefficients must match:

k “ 1 (2)

m` kpa´ bq “ 2pa´ bq (3)

mpa´ bq ` kpb´ 2q “ a (4)

mpb´ 2q “ 15´ 2pa´ bq (5)

From (4), (2), and (3) we obtain:

»

—

—

–

m “ a´ b;

a “ pa´ bq2 ` b´ 2

fi

ffi

ffi

fl

(6)

Also, from m “ a´ b and (5) we get:

pa´ bqpb´ 2q “ 15´ 2pa´ bq;

pa´ bqpb´ 2q ` 2pa´ bq “ 15;

21

pa´ bq rpb´ 2q ` 2s “ 15;

»

—

—

–

bpa´ bq “ 15;

b ‰ 0, a ‰ b, a´ b “ 15b

fi

ffi

ffi

fl

(7)

Using (7) and (6) we see that,

b` 15b “

`

15b

˘2` b´ 2 (8)

Multiplying both sides of (8) by b2 yields,

b3 ` 15b “ p15q2 ` b2pb´ 2q;

b3 ` 15b “ p15q2 ` b3 ´ 2b2;

2b2 ` 15b´ p15q2 “ 0 (9)

The discriminant of the quadratic equation (in b) in (9), is D “ p15q2 ` 4p2qp15q2 “ p15q2 ˚ 32.

The two real roots of equation (9) are;

b “ ´15˘p15qp3q4 “ 30

4 ,´604 ; b “ 15

2 ,´15

When b “ 152 , we get (from (7)) a “ 15

2 ` 2 “ 192 .

Also, m “ a´ b “ 192 ´

152 “

42 “ 2.

So the linear polynomial kx ` m is x ` 2 in this case and the quadratic polynomial gpxq is

gpxq “ x2 ` 2x` 112 ; which has two complex roots:

´2`3i?2

2 and ´p2`3i?2q

2 .

Obviously, since fpxq “ pkx `mqgpxq; the three roots of fpxq; are the two roots of gpxq and the

real number ´mk . We have the following:

One solution to this problem is a “ 192 , b “ 15

2 .

The roots of fpxq in this case are: ´2, ´2`3i?2

2 ,´2`3i?2

2

22

The other solution comes from b “ ´15. Then, a “ b ` 15b “ ´15 ´ 1 “ ´16; and m “ a ´ b “

´16` 15 “ ´1 And so, ´mk “ ´

´11 “ 1.

In this case, gpxq “ x2 ` pa´ bqx` b´ 2; gpxq “ x2 ´ x´ 17; which has the two real roots 1˘?69

2 .

We have the following:

The other solution to this problem occurs when a “ ´16 and b “ ´15.

The three roots of fpxq in this case are the real numbers 1, 1`?69

2 , 1´?69

2

23

Problem 6 The given equation is:

x3 ´ pk ` 1qx` 2k “ 0 where k P Z (1)

Since Equation (1) has integer coefficients, it follows from the Rational Root Theorem for poly-

nomial functions with integer coefficients; that if r is a rational root of Equation (1), then r “ dq ;

where d is a divisor of the constant 2k; and q is a divisor of the leading coefficient, which is 1. Thus,

q “ 1 or ´1. Thus, r “ ˘d, r P Z, which shows that r must be a divisor of 2k:

(r P Z and 2k “ r ˚m, for some m P Z (2)

And since r satisfies Equation (1), we also have:

r3 ´ pk ` 1qr ` 2k “ 0;

ÐÑ

2r3 ´ 2pk ` 1qr ` 4k “ 0 (3)

By (2) and (3) we have:

2r3 ´ r2m´ 2r ` 2rm “ 0

ÐÑ

rp2r2 ´ rm´ 2` 2mq “ 0 (4)

24

It is clear from (4) and (3) that r “ 0 if and only if k “ 0. When k “ 0, Equation (1) sim-

ply becomes x3 ´ x “ 0; which has, in fact, three rational roots: the integers ´1, 0, and 1.

Next, assume that:

r ‰ 0 (5)


2pr2 ´ 1q ´mpr ´ 1q “ 0;

ÐÑ

2pr ´ 1qpr ` 1q ´mpr ´ 1q “ 0;

ÐÑ

pr ´ 1q r2pr ` 1q ´ms “ 0 (6)

If r “ 1 in (6), then from (3) we get;

2´ 2k ´ 2` 4k “ 0;

k “ 0

Once again, Equation (1) becomes x3 ´ x “ 0; which has three rational roots, ´1, 0, and 1.

Next, suppose that r ‰ 1. Then by (6);

r ‰ 1 and m “ 2pr ` 1q (7)

25

From (7) and (2) we further get that:

2k “ 2rpr ` 1q;

k “ rpr ` 1q (8)

By (3) and (8) we obtain:

r3 ´ rrpr ` 1q ` 1s r ` 2rpr ` 1q “ 0; and since r ‰ 0 (by (5));

ÐÑ

r2 ´ rpr ` 1q ´ 1` 2pr ` 1q “ 0

ÐÑ

r2 ´ r2 ´ r ´ 1` 2r ` 2 “ 0

ÐÑ

r “ ´1

Again, r “ ´1 yields k “ 0 by (8).

Conclusion: This problem has a unique solution:

k “ 0

26

Problem 7

Consider the family of polynomial functions:

fpxq “ x4 ´ pp` 3qx2 ` p2, where p is a real parameter (1)

By inspection we see that 0 is one of the roots of fpxq, precisely when p “ 0. In that case, the other

two roots are?

3 and ´?

3. We have the following:

For p “ 0, the polynomial function has zero as a root of multiplicity 2;

and the other two roots are the real numbers?

3 and ´?

3.

Next, assume that p ‰ 0 (2).

The four roots of fpxq are found from the two roots r1 and r2 of the quadratic trinomial,

gptq “ t2 ´ pp` 3qt` p2 (2a)

By solving for x, the two equations:

x2 “ r1 and x2 “ r2 (3)

we can determine the four roots of fpxq. It is clear from (3), that the four roots of fpxq will be real

if, and only if, the numbers r1 and r2 are non-negative real numbers. But neither r1 nor r2 can be

zero, by virtue of (2). Thus, the four roots of fpxq will be real numbers, precisely when;

r1 ą 0 and r2 ą 0 (4)

Consider (4) and (2a). The two roots r1 and r2 of gptq, will be positive real numbers if, and only if,

one of two sets of conditions is satisfied:

Condition 1 (C1):

D “ pp` 3q2 ´ 4p2 “ 0

This is the case of a double positive real root;

r1 “ r2 “p`32

27

Condition 2 (C2):

D “ pp` 3q2 ´ 4p2 ą 0; where p ‰ 0

and (sum) r1 ` r2 “p`31 “ p` 3 ą 0

and (product) r1 ˚ r2 “ p2 ą 0, already satisfied by (2)

This is the case of two distinct positive real roots

We factor the discriminant D of the trinomial gptq into a product of two linear factors:

D “ pp` 3q2 ´ 4p2 “ rpp` 3q ´ 2ps rpp` 3q ` 2ps;

D “ p3´ pqp3p` 3q “ 3pp` 1qp3´ pq (5)

It is clear from (5) that D “ 0 Ø p “ ´1 or 3. So conditions C1 occur in exactly two instances:

p “ ´1, with double root of gptq being r1 “ r2 “ 1 and for p “ 3; r1 “ r2 “ 3.

From (5), we see that for conditions C2 to be satisfied; we must have:

»

—

—

–

3pp` 1qp3´ pq ą 0

p` 3 ą 0; p ‰ 0

fi

ffi

ffi

fl

Ø

»

—

—

–

´1 ă p ă 3

p ą ´3; p ‰ 0

fi

ffi

ffi

fl

;

which is equivalent to ´1 ă p ă 3, with p ‰ 0. Using (3), we obtain the four distinct real

roots. In this case:

x “ ˘?r1,˘

?r2.

28

Summary of Results

The biquadratic polynomial function in (1) will have four real roots in precisely the following

four cases:

1. p “ 0. The four real roots being 0(multiplicity 2),?

3, and ´?

3.

2. p “ 3. The four real roots being?

3 and ´?

3 (each having multiplicity 2).

3. p “ ´1. The four real roots being 1 and ´1 (each having multiplicity 2).

4. ´1 ă p ă 3 and p ‰ 0. The four real roots are the distinct real numbers?r1, ´

?r1,

?r2,

and ´?r2, where r1 “

p`3`?

3pp`1qp3´pq

2 , r2 “p`3´

?3pp`1qp3´pq

2 .

29

Problem 8 We have:¨

˚

˚

˝

logaploga xq “ loga2 rloga2 xs

a ą 0, a ‰ 1

˛

‹

‹

‚

(1)

The given equation (1) is equivalent to:

pa2qlogaploga xq “ pa2qloga2 ploga2 xq;

ÐÑ

“

alogaploga xq‰2“ loga2 x;

ÐÑ

ploga xq2 “ loga2 x (2)

Equation (2) is equivalent to:

pa2qploga xq2 “ pa2qloga2 x

ÐÑ

”

aploga xq2ı2“ x;

ÐÑ

`

aloga x˘2 loga x

“ x;

ÐÑ

x2 loga x “ x “ x1 (3)

Equation (3) requires that the exponents must match:

2 loga x “ 1 Ø loga x “12 ; or equivalently,

x “ a12 “

?a;

Therefore, the unique solution to the given equation is:

x “?a

30

Problem 9 We have the 3-variable system:

¨

˚

˚

˝

1x `

1y `

1z “ 1

xyz “ 2; and x ă y ă 0 ă x

˛

‹

‹

‚

(1)

System (1) is equivalent to the system:

¨

˚

˚

˝

xy ` yz ` zx “ xyz,

xyz “ 2 and z ă y ă 0 ă x

˛

‹

‹

‚

(2)

The first and second equations in (2) give xpy ` zq “ 2 ´ yz; y ` z “ 1x p2´ yzq; and since

yz “ 2x , we further obtain:

y ` z “ 2px´1qx2 (3)

Considering the inequalities z ă y ă 0 ă x in (2); it becomes clear that (3) requires that since

y ` z ă 0; we must have:

0 ă x ă 1 (4)

Combining (2), (3), and (4); we see that the system in (2) is transformed in the equivalent system:

¨

˚

˚

˚

˚

˚

˚

˝

y ` z “ 2pt´1qt2

yz “ 2t , x “ t

And with z ă y ă 0 ă t ă 1

˛

‹

‹

‹

‹

‹

‹

‚

(5)

According to (5), the negative real numbers z and y are the two roots of the quadratic trino-

mial in w;

fpwq “ w2 ´

´

2pt´1qt2

¯

w ` 2t (6)

31

The sum of the two roots is 2pt´1qt2

ă 0 and the product is 2t ą 0. The discriminant D of this

trinomial is given by:

D “

”

2pt´1qt2

ı2´ 4

`

2t

˘

“4pt´1q2

t4´ 8

t ;

ÐÑ

D “4rpt´1q2´2t3s

t4“

4r´2t3`t2´2t`1st4

(7)

Note that the number 12 is a real root of the cubic polynomial ´2t3 ` t2 ´ 2t ` 1. If we per-

form synthetic division of the cubic polynomial with t ´ 12 ; we obtain a quotient of ´2t2 ´ 2 with

remainder zero. Further, we have the factorization:

´2t3 ` t2 ´ 2t` 1 “ p´2t` 1qpt2 ` 1q (8)

Note: We could have factored just as easily using group factoring as follows:

“

´2t3 ` t2 ´ 2t` 1 “ ´2tpt2 ` 1q ` pt2 ` 1q “ p´2t` 1qpt2 ` 1q‰

Thus, from (7) and (8) we obtain:

D “4p´2t`1qpt2`1q

t4(9)

Since the trinomial in (6) has two distinct real roots, (z ă y ă 0 by (5)); the discriminant D

must be positive; which by (9) implies that ´2t ` 1 ą 0; t ă 12 ; which when combined with (5)

gives,

0 ă t ă 12 (10)

32

The two real negative roots z and y of the trinomial fpwq in (6) are given by (since z ă y):

z “ 12

”

2pt´1qt2

´?Dı

, y “ 12

”

2pt´1qt2

`?Dı

(11)

From (5), (10), (11), we arrive at the following conclusion:

All the solutions to the system in (1) are given by:

x “ t, 0 ă t ă 12

z “t´1´

?p´2t`1qpt2`1q

t2

y “t´1`

?p´2t`1qpt2`1q

t2

33

Problem 10 The two trinomials are:

t1pxq “ x2 ´ px` q and t2pxq “ x2 ` qx´ p (1)

If r is the one root that t1pxq and t2pxq have in common, then:

¨

˚

˚

˝

r2 ´ pr ` q “ 0; and

r2 ` qr ´ p “ 0

˛

‹

‹

‚

(2)

From (2) it follows that:

r2 “ pr ´ q “ p´ qr

ÐÑ

pr ´ p´ q ` qr “ 0

ÐÑ

pp` qqpr ´ 1q “ 0 (3)

According to equation (3); we must have:

r “ 1 or p` q “ 0 (4)

If p` q “ 0; p “ ´q, then by (1) we see that, t1pxq “ x2´ px´ p and t2pxq “ x2´ px´ p. The two

trinomials would be identical, contradicting the hypothesis that the two trinomials have only one

root in common.

Thus, (4) implies that:

r “ 1 and p` q ‰ 0 (5)

Going back to (2) with r “ 1, we find that:

p´ q “ 1

ÐÑ

p “ 1` q (6)

34


t1pxq “ x2 ´ pq ` 1qx` q and t2pxq “ x2 ` qx´ pq ` 1q (7)

Since r “ 1 is the common root, the other root R1 of t1pxq is given by:

R1 ` r “ q ` 1

ÐÑ

R1 ` 1 “ q ` 1

ÐÑ

R1 “ q (8)

And the other root R2 of t2pxq is given by,

R2 ` r “ ´q

ÐÑ

R2 ` 1 “ ´q

ÐÑ

R2 “ ´p1` qq (9)

So the trinomial t3pxq that has the roots R1 and R2, has indeed two rational roots, by (8) and (9):

The rational numbers q and ´p1` qq and:

t3pxq “ px´ qqrx´ p´p1` qqqs;

ÐÑ

t3pxq “ px´ qqrx` p1` qqs;

ÐÑ

t3pxq “ x2 ` x´ qp1` qq.

35

Problem 11: We have the system,

»

—

—

—

—

—

—

–

x` 2y ´ 3z “ 4

3x´ y ` 5z “ 2

4x` y ` pc2 ´ 14qz “ c` 2

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

We use the method of Gaussian elimination, in order to reduce the augmented matrix of this

system, to a row-echelon matrix form.

Notation:

1) The notation kRi `Rj will mean that we replace the jth row by the combination kRi `Rj(the

ith row multiplied with k plus the jth row).

2) The notation kRi will mean that we replace the ith row by k multiplied with the ith row.

We have:»

—

—

—

—

—

—

–

1 2 ´ 3 | 4

3 ´ 1 5 | 2

4 1 c2 ´ 14 |c` 2

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

´3R1 `R2

´4R1 `R3

ÐÑ

»

—

—

—

—

—

—

–

1 2 ´ 3 | 4

0 ´ 7 14 | ´ 10

0 ´ 7 c2 ´ 2 | c´ 14

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

´R2 `R3

ÐÑ

36

»

—

—

—

—

—

—

–

1 2 ´ 3 | 4

0 ´ 7 14 | ´ 10

0 0 c2 ´ 16 | c´ 4

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

;

which is row-echelon form. Therefore, the given system is equivalent to the system:

»

—

—

—

—

—

—

–

x` 2y ´ 3z “ 4

´7y ` 14z “ ´10

pc2 ´ 16qz “ c´ 14

fi

ffi

ffi

ffi

ffi

ffi

ffi

fl

p1q

p2q

p3q

We focus on Equation (3): pc´ 4qpc` 4qz “ c´ 4. If c “ ´4, the last equation becomes 0 ˚ z “ ´8,

which cannot be satisfied for any value of the variable z. It is clear, then, that the system has no

solutions for c “ ´4.

If c “ 4, then (3) becomes 0 ˚ z “ 0, which is satisfied for any real value of z. However, putting

z “ t P R; we get, from (2), y “ 2t ` 107 ; and from Equation (1), x “ ´t ` 8

7 . So, in this case, we

have infinitely many solutions.

Finally, if c ‰ ´4 and c ‰ 4, we have a unique solution:

From (3), z “ 1c`4 ; and by back substitution, (2) and (3) yield y “ 10c`54

7pc`4q and x “ 8c`47pc`4q .

Summary

1. If c ‰ ´4 and c ‰ 4, the given system has the unique solution: px, y, zq “´

8c`47pc`4q ,

10c`547pc`4q ,

1c`4

¯

2. If c “ ´4, the system has no solutions.

3. If c “ 4, the system has infinitely many solutions, which can be parametrically described by

the formulas: x “ ´t` 87 , y “ 2t` 10

7 , z “ t; t P R

37

Problem 12

Let x and y be the two dimensions (or side lengths) of such a rectangle. Then, the semi-perimeter

is equal to 2px`yq2 “ x “ y, and each diagonal has length

a

x2 ` y2. According to the hypothesis of

the problem, the positive integers x and y must satisfy the equation:

(Area) xy “ 4”

x` y `a

x2 ` y2ı

;

ÐÑ»

—

—

–

xy ´ 4px` yq “ 4a

x2 ` y2

and with xy ´ 4px` yq ą 0; x, y are positive integers

fi

ffi

ffi

fl

(1)

Squaring both sides of the equation in (1):

pxyq2 ´ 8xypx` yq ` 16px` yq2 “ 16px2 ` y2q;

ÐÑ

pxyq2 ´ 8xypx` yq ` 16px2 ` y2q ` 32xy “ 16px2 ` y2q;

ÐÑ

pxyq2 ´ 8xypx` yq ` 32xy “ 0;

and since xy ą 0, the last equation is equivalent to:

xy ´ 8px` yq ` 32 “ 0;

xy ´ 8px` yq ` 64 “ 32;

px´ 8qpy ´ 8q “ 32 (2)

Considering the divisors of 32, we see that there are exactly twelve possibilities:

1. x ´ 8 “ 1 and y ´ 8 “ 32; which gives px, yq “ p9, 40q which does satisfy the inequality

condition in (1).

38

2. x ´ 8 “ ´1 and y ´ 8 “ ´32; which gives px, yq “ p7,´24q, which is not a solution, since

´24 ă 0.

3. x´ 8 “ 2 and y ´ 8 “ 16; which gives px, yq “ p10, 24q which satisfies (1).

4. x´ 8 “ ´2 and y ´ 8 “ ´16; which gives px, yq “ p6,´8q, which is not a solution.


6. x´ 8 “ ´4 and y ´ 8 “ ´8; which gives px, yq “ p4, 0q which is not a solution.


8. x´ 8 “ ´8 and y ´ 8 “ ´4; which gives px, yq “ p0, 4q which is not a solution.


10. x´ 8 “ ´16 and y ´ 8 “ ´2; which gives px, yq “ p´8, 6q which is not a solution.


12. x´ 8 “ ´32 and y ´ 8 “ ´1; which gives px, yq “ p´24, 7q, which is not a solution.

Obviously, because of the obvious symmetry, possibilities 7-12 could be safely omitted.

Conclusion

There are exactly three groups of rectangles that satisfy this problem’s property:

1. The rectangles of length dimensions 9 and 40.



39

Problem 13 We have the system:

»

—

—

—

—

—

—

—

—

—

—

–

x` y “ 100

xy “ z2

1x `

1y “

2w

And with w, x, y, z being positive integers

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÐÑ

»

—

—

—

—

—

—

—

—

—

—

–

x` y “ 100

xy “ z2

wpx` yq “ 2xy;

w, x, y, z are positive integers

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÐÝ

»

—

—

—

—

—

—

—

—

—

—

–

x` y “ 100

xy “ z2

50w “ z2;

w, x, y, z are positive integers

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

p1q

p2q

p3q

Looking at Equation (3), we see that 10 must divide z:

z “ 10u, u P Z` (4)

From (3) and (4), we obtain:

w “ 2u2 (5)

40

Also, from (2) and (4), we have:

xy “ 100u2 (6)

According to (1) and (6), the positive integers x and y are the roots of the quadratic equation in t:

t2 ´ 100t` 100u2 “ 0 (7)

The discriminant D of Equation (7) is given by:

D “ p´100q2 ´ 4p100qu2 “ 100“

100´ 4u2‰

“ 400“

25´ u2‰

(8)

The two roots or zeros of Equation (7) are:

100`20?25´u2

2 and 100´20?25´u2

2 ; or

`

50` 10?

25´ u2 and 50´ 10?

25´ u2˘

(9)

These two roots will be integers if, and only if, 25´ u2 is a perfect square, which occurs if u “ 3 or

u “ 4 (recall from (4) that u is a positive integer).

If u “ 3, the two roots in (9) are 90 and 10.

If u “ 4, (9) gives 80 and 20.

Since x and y are the two roots, we have x “ 10, 20, 80, or 90 and correspondingly y “ 90, 80, 20,

or 10. So for u “ 3, we get z “ 30, w “ 32 for u “ 4, z “ 40, and w “ 32 from (4) and (5).

Conclusion

This system has exactly four solutions:

px, y, w, zq =

»

—

—

—

—

—

—

—

—

—

—

–

p10, 90, 18, 30q

p90, 10, 18, 30q

p20, 80, 32, 40q

p80, 20, 32, 40q

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

41

Problem 14 The squares modulo 9 are 0, 1, 4, and 7. Thus, if a is any integer:

a2 ” 0, 1, 4, or 7 (mod 9) (1)

Using (1), we can determine all possible values, modulo 9, of the sum of two perfect or integer

squares as follows.

If x1 and x2 are integers, then:

x21 ` x22 ”

»

—

—

—

—

—

—

—

—

—

—

–

0, 2, 8, 14,

1, 2, 5, 8,

4, 5, 0, 2,

7, 8, 11, 14

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

;

x21 ` x22 ” 0, 1, 2, 4, 5, 7, or 8 (mod 9) (2)

Therefore, the sum of two perfect squares cannot equal, according to (2); to 3 or 6 (mod 9) (3)

On the other hand, since 10 ” 1(mod 9); a natural number with only 2, one 1 and the rest of

the digits all zero will be congruent to:

2` 1 ” 3(mod 9) (4)

This follows easily from the base 10 representation of an integer. Thus, it follows (3) and (4),

that such an integer cannot equal the sum of two perfect squares.

Next, the cubes are examined. If a P Z; a3 ” 0, 1, or 8 (mod 9). It follows that x31 ` x32 ” 0, 1, 2 or

8(mod 9). And so, x31 ` x32 ı 3 (mod 9), which again proves that such an integer cannot equal the

sum of two perfect cubes.

42

References

1. Crux Mathematicorum with Mathematical Mayhem Vol. 38, No 7, September 2012 issue

Olympiad Corner OC 91, page 267

2. Howard Anton, Elementary Linear Algebra, Applications Version, Seventh Edition, 1994.

Published by John Wiley & Sons, ISBN # 0-471-58741-9, Exercise 17, on page 22,

in Section 1.2

3. Konstantine Zelator, Nine Classical Algebra Problems from a modern Greek mathematics book,

published in www.academia.edu, Publication Date: July 12, 2013.

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