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Which steps I can follow to find the integer solutions for the equation $x^2 + y^2 = 208$?

Mateus Coutinho Marim
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3 Answers3

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A positive integer is the sum of two squares iff primes $\equiv 3\pmod 4$ (following the Fermat's theorem on sums of two squares) of its factorization are raised to even exponents.

$$208=2^4\cdot 13$$

So, $208$ meets the condition.

Now, $13=3^2+2^2$, so $208=4^2(3^2+2^2)=12^2+8^2$

ajotatxe
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  • Thanks, can you tell me what can I study to learn how to do this? I did not studied this yet. – Mateus Coutinho Marim Jul 13 '15 at 17:56
  • http://math.stackexchange.com/questions/1341756/how-do-i-find-two-integers-x-and-y-whose-values-satisfy-the-expression/1341774#1341774 – Batominovski Jul 13 '15 at 18:12
  • I studied this in my first course of my grade in Maths. – ajotatxe Jul 13 '15 at 19:30
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    @MateusCoutinhoMarim, many number theory textbooks have these results. One book is Hardy and Wright's Introduction to the Theory of Numbers. That may be a bit dense though. There is also Strayer's Elementary Number Theory, which is fairly easy to read. – Joel Jul 14 '15 at 18:21
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Since $208$ is a relatively small number, it is not too hard to check $208-x^2$ for all numbers $0\leq x\leq 14$ to see if the answer is a square number. The negatives of whatever solutions you find are also solutions.

Plutoro
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This represents a circle centered at the origin of radius $\sqrt{208}$. You only need to check integer $x$-coordinates between $-14$ and $14$.

Michael Hardy
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John McGee
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