1980 AHSME Problems/Problem 29

Problem

How many ordered triples (x,y,z) of integers satisfy the system of equations below?

$\begin{array}{l} x^2-3xy+2y^2-z^2=31 \\ -x^2+6yz+2z^2=44 \\ x^2+xy+8z^2=100\\ \end{array}$

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \\ \text{(D)}\ \text{a finite number greater than 2}\qquad\\ \text{(E)}\ \text{infinitely many}$

Solution 1

Sum of three equations,

$x^2-2xy+2y^2+6yz+9z^2 = (x-y)^2+(y+3z)^2 = 175$

(x,y,z) are integers, ie. $175 = a^2 + b^2$ ,

$a^2$ : 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169 $b^2$ : 174, 171, 166, 159, 150, 139, 126, 111, 94, 75, 54, 31, 6

so there is NO solution

Wwei.yu (talk) 22:09, 28 March 2020 (EDT)Wei

Solution 2 (Eliminate Cross Terms)

First, to eliminate the $xy$ term, multiply both sides of the third equation by $3$ to get $3x^2 + 3xy + 24z^2 = 300$ , and then add it to the first equation to get $4x^2 + 2yz + 23z^2 = 331$ .

Next, to eliminate the $yz$ term, multiply both sides of the new equation by $-3$ to get $-12x^2 - 6yz - 69z^2 = -993$ , and then add it to the second equation to get $-13x^2 - 67z^2 = -949$ .

This equation can be rearranged to get $67z^2 = 949-13x^2 = 13(73-x^2)$ . Since $67$ is not divisible by $13$ , $z$ must be divisible by $13$ and there exists an integer $n$ such that $z = 13n$ . Substituting $z = 13n$ and dividing both sides of this equation by $13$ yields $871n^2 = 73 - x^2$ . So $871n^2 \leq 73$ , which is only true if $n = 0$ . But $n = 0$ would require $x^2 = 73$ , which is impossible since $73$ is not a perfect square. Therefore, this equation (and also this system) has $0$ integer solutions. $\fbox{(A)}$

-j314andrews, based on solution by Farenhajt

Solution 3 (Modular Arithmetic)

Reducing the second equation mod $2$ yields $x^2 \equiv 0\ (\mathrm{mod}\ 2)$ . Therefore, $x \equiv 0\ (\mathrm{mod}\ 2)$ and $x$ is even.

Reducing the first equation mod $2$ and substituting yields $z^2 \equiv 1\ (\mathrm{mod}\ 2)$ . Therefore, $z \equiv 1\ (\mathrm{mod}\ 2)$ and $z$ is odd.

Since $x$ is even, $x^2 \equiv 0\ (\mathrm{mod}\ 4)$ , and since $z$ is odd, $z^2 \equiv 1\ (\mathrm{mod}\ 4)$ . Also $6z \equiv 2z \equiv 2\ (\mathrm{mod}\ 4)$ . Reducing the second equation mod $4$ and substituting yields $2y + 2 \equiv 0\ (\mathrm{mod}\ 4)$ , so $y$ is odd.

Reducing the third equation mod $4$ and substituting yields $xy \equiv 0\ (\mathrm{mod}\ 4)$ . Since $y$ is odd, $x \equiv 0\ (\mathrm{mod}\ 4)$ .

Since $y$ is odd, $y^2 \equiv 1\ (\mathrm{mod}\ 4)$ . Reducing the first equation mod $4$ and substituting yields $1 \equiv 3\ (\mathrm{mod}\ 4)$ , which is impossible.

Therefore, this system has $0$ integer solutions. $\fbox{(A)}$

-j314andrews

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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1980 AHSME Problems/Problem 29

Contents

Problem

Solution 1

Solution 2 (Eliminate Cross Terms)

Solution 3 (Modular Arithmetic)

See also