1980 AHSME Problems/Problem 29

Problem

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

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

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

Solution 1 (Sum of Squares)

We are considering the sum of three equations:

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

Since $x, y, z$ are integers, this implies that $175$ can be written as the sum of two squares, i.e.,

$175 = a^2 + b^2,$

where $a$ and $b$ are integers.

However, $175 \equiv 3\ (\mathrm{mod}\ 4)$ , but $a^2$ and $b^2$ must each be congruent to either $0$ or $1 (\mathrm{mod}\ 4)$ . So this equation, and therefore this system, has $0$ integer solutions. $\fbox{(A)}$

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

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 (Sum of Squares)

Solution 2 (Eliminate Cross Terms)

Solution 3 (Modular Arithmetic)

See Also