**Problem**: Prove that there are no positive integer solutions to $ x^2+y^2 = 3z^2 $.

**Solution**: Let's use our favorite tool - mods. Suppose there exists a solution $ (x,y,z) $; if they share a common factor, we can divide it out, so we can assume WLOG that they do not.

Taking modulo $ 3 $, we get

$ x^2+y^2 \equiv 0 \pmod{3} $.

We can easily see that this requires both $ x $ and $ y $ to be divisible by $ 3 $. Let $ x = 3x_0 $ and $ y = 3y_0 $. Using our original equation,

$ 9x_0^2+9y_0^2 = 3z^2 $

$ 3(x_0^2+y_0^2) = z^2 $.

Modulo $ 3 $ again,

$ z^2 \equiv 0 \pmod{3} \Rightarrow z \equiv 0 \pmod{3} $.

But then $ 3 $ divides each of $ x,y,z $, contradicting our assumption that they do not have a common factor. Hence no solution exists. QED.

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Comment: More elementary number theory and work with mods; this uses the idea of infinite descent implicitly as well (i.e. we can show that $ x, y, z $ are infinitely divisible by $ 3 $). Diophantine equations are often simplified greatly through mods.

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Practice Problem: What positive integers can we replace $ 3 $ with such that the problem statement still holds?

Any number whose prime factorization contains a prime congruent to 3 mod 4, and does not contain the square of that prime.

ReplyDeleteSorry, I should be a bit more general than that. Precisely the numbers whose prime factorization contains a prime congruent to 3 mod 4 raised to an odd power. This is easy to deduce using the arithmetic of the Gaussian integers.

ReplyDeleteI believe that simplifies to all numbers that are not the sum of two squares :)

ReplyDeleteThis kinda reminds me of the question, prove that there are no rational solutions to x^2 + y^2 = 3. Fun stuff.

ReplyDelete