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Given a number say $x$, how to check if it can become hypotenuse of right angle triangle and other sides must be integer

For example:

$5$ it can be hypotenuse as its other sides $3$ & $4$ are integers.

$13$ it can also hypotenuse as its other sides $12$ & $5$ are integers.

$12$ can't be hypotenuse because other two sides can't be integers.

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An integer number $N>0$ can be the hypothenuse of a right angled triangle with integer side lengths if and only if it has a prime factor $p \equiv 1 (4)$ (i.e. $p$ is of the form $4k+1$ with integer $k$).

The proof outline is as follows:

  • If $N^2 = a^2+b^2$ then $(kN)^2 = (ka)^2+(kb)^2$.
  • Any prime $p \equiv 1 (4)$ can be represented as $x^2+y^2$ with positive integers $x,y$ (Fermat's theorem on sums of two squares). And then $p^2 = (x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2$
  • All other numbers only have the trivial representation $N^2 = N^2+0^2$ (see e.g. here), which does not form a triangle.
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  • $\begingroup$ Thank You so much. Simple, Consise and to the point. $\endgroup$ – DollarAkshay Oct 7 '15 at 18:32
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It is known that any primitive pythagorean triple $a^2 + b^2 = c^2$ is on the form $$ (a, b, c) = (2uv, u^2 - v^2, u^2 + v^2) $$for some natural numbers $u>v$, so you basically have to check whether your hypotenuse can be written as the sum of two squares. To confirm your examples, $5 = 4 + 1$, and $13 = 9 + 4$, while $12$ is not the sum of two squares (you can't add your way to $12$ by using only two numbers from $\{1, 4, 9\}$).

If course, it's possible that your number could be the hypotenuse in some non-primitive triple. This happens if it has a divisor that can be written as the sum of two squares. For instance, if at least one of its prime factors are congruent to $1$ modulo $4$.

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    $\begingroup$ Your interesting observation is false: $15$ can't be written as the sum of two squares, but $225=12^2+9^2$ can. (It's true with "only if" instead of "if and only if".) $\endgroup$ – TonyK Oct 2 '15 at 15:42

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