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I put together a VBA script to find me all of the Pythagorean Triangles from 3,4,5 all the way to 105, 608, 617. Then I arranged them by increasing hypotenuse $c$, and identified all of the triangles where $c$ was a prime number. I then subtracted each $c$ value from the preceeding prime $c$ and found that in every instance the difference between the two values was divisible by 4.

Example 1:

$$12^2+35^2=37^2$$ $$9^2+40^2=41^2$$

and $41-37 \equiv 0 \pmod 4$.

Example 2:

$$9^2+40^2=41^2$$ $$29^2+45^2=53^2$$

and $53-41 \equiv 0 \pmod 4$. And so forth.

Is there a way to prove whether or not this holds true for all prime numbers that can be represented by the square root of the sum of two square integers?

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The reason is that the hypotenuse is of the form $$ c = \lambda (u^{2} + v^{2}), $$ where $u, v > 0$ (are coprime and) have different parity, that is, one of them is even, and the other is odd.

If such a $c$ is prime, then $\lambda = 1$, and then $u^{2} + u^{2} \equiv 1 \pmod{4}$.

Hence the difference of two such $c$'s is $\equiv 0 \pmod{4}$.

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You can represent every primitive Pythagorean Triple with $\{M^2-N^2,2MN,M^2+N^2\}$, where the last number represents the hypotenuse. For $M^2+N^2$ to be prime, $M^2+N^2\equiv1\bmod4$ or $M^2+N^2\equiv3\bmod4$. The second case can be ignored because $M^2\equiv0\bmod4$ or $M^2\equiv1\bmod4$ for all integers, so $M^2+N^2\equiv0,1,\text{or }2\bmod4$. This means that all prime hypotenuses must be one more than a multiple of four, so their difference is always going to be divisible by four.

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  • $\begingroup$ @AndreasCaranti: He means every primitive Pythagorean triple. That is, no common factor. $\endgroup$ – Tito Piezas III Jan 13 '17 at 14:58
  • $\begingroup$ @AndreasCaranti Thank you. $\endgroup$ – AlgorithmsX Jan 13 '17 at 15:00

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