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1980 AMC 12 Problem 29

Attempt Problem 29 of the 1980 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1980 AMC 12 solutions, or check the answer key.

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29.

How many ordered triples (x,y,z)(x,y,z) of integers satisfy the system of equations below? x2−3xy+2y2−z2=31,−x2+6yz+2z2=44,x2+xy+8z2=100. \begin{aligned} x^2-3xy+2y^2-z^2&=31,\\ -x^2+6yz+2z^2&=44,\\ x^2+xy+8z^2&=100. \end{aligned}

00

11

22

a finite number greater than two

infinitely many

Answer: A
Concepts:Diophantine Equationmodular arithmeticperfect square
Difficulty rating: 2100
Small Hint:

Add all three equations before trying to solve for the variables

Big Hint:

Rewrite the resulting quadratic form as a sum of two squares and reduce modulo 44

Solution:

Adding the three equations gives x2−2xy+2y2+6yz+9z2=175, \begin{aligned} x^2-2xy+2y^2 &{}+6yz+9z^2\\ &=175, \end{aligned} or (x−y)2+(y+3z)2=175. (x-y)^2+(y+3z)^2=175. A square is congruent to 00 or 1(mod4),1\pmod4, so a sum of two squares cannot be congruent to 3(mod4).3\pmod4. But 175≡3(mod4),175\equiv3\pmod4, a contradiction. Thus there are no integer triples.

Therefore, the correct answer is A.

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