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I wish to prove that for any coprime (and positive) Pythagorean triple $(x,y,z)$, where $x$ is even, $x+z$ is always a square number.

By inspection, this appears true; $(3,4,5)$ gives $9$, $(5,12,13)$ gives $25$, $(8,15,17)$ gives $25$, $(7,24,25)$ gives $49$ et cetera. Yet I am stuck on the proof.

I do not know the original source of the problem.

Davood
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Heinrich Wagner
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    All Pythagorean triples can be generated by choosing integer values for $a$ and $b$ and substituting them into \begin{eqnarray} x=a^2-b^2 \ y=2ab \ z =a^2+b^2 \end{eqnarray} – Donald Splutterwit Aug 24 '17 at 20:40
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    I think this is only true for primitive triples. (12, 16, 20) is a counterexample. It comes out of Euclid's formula. https://en.wikipedia.org/wiki/Pythagorean_triple – Shuri2060 Aug 24 '17 at 20:40
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    OP did say "coprime", which is the same as primitive. – marty cohen Aug 24 '17 at 20:45
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    @martycohen Pretty sure that 'coprime' wasn't there before... – Bram28 Aug 24 '17 at 20:47
  • @Heinrich Wagner See each of the following: https://math.stackexchange.com/questions/395543/formulas-for-calculating-pythagorean- triples https://math.stackexchange.com/questions/1826816/how-to-prove-pythagorean-triple-formula – Davood Aug 24 '17 at 20:49
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    @Bram28: You may be right. I find it annoying when a question is edited to fix an error and the change is not mentioned. – marty cohen Aug 24 '17 at 21:12
  • It doesn't matter if they are coprime or not.

    Given $$A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$$

    $$C-B=m^2-2mn+n^2=(m-n)^2$$ $$B+C=m^2+2mn+n^2=(m+n)^2$$

     – poetasis Jul 07 '19 at 16:54
  • @Heinrich Wagner I do not believe this question should have been closed because the indicated link does not answer the question of whether the sum of the even side and the hypotenuse is always a perfect square.

    An alternate formula generates all Pythagorean triples where the greatest common divisor of $A,B,C$ is an odd square, includes all primitives and always generates side-$B$ even. $$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$ If we add $B$ and $C$, we get

    $$[2(2n-1)k+2k^2]+[(2n-1)^2+2(2n-1)k+2k^2]\quad =2(2n-1)^2+4(2n-1)k+k^2\quad =(2(2n-1)+k)^2$$

     – poetasis Mar 14 '20 at 18:09

2 Answers2

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For every Pythagorean triple $(x,y,z)$, with $d=\gcd(x,y,z)$;
there are coprime integers $a,b$ with different pairity, such that:
$$x=d(a^2-b^2) \ \text{and} \ y=d(2ab) \ \text{and} \ z=d(a^2+b^2); $$

Now notice that the coprime condition implies $d=1$; so we have the following:

$$(2ab)+(a^2+b^2)=(a+b)^2.$$

Davood
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Note that $(m^2+n^2)^2=(2mn)^2+(m^2-n^2)$. Since $2mn$ is always even, we have $m^2+n^2+2mn=(m+n)^2$.

As @shuri pointed out, in $(3,4,5)$, $4+5=9$ which is a perfect square. But when you do not include your coprime condition, we can multiply all the elements of triplet by an even number say $2,4$ etc to get $(x,y,z)$ of same parity and then one pair give us a perfect square $m^2+n^2+2mn$ while other one give us twice of a perfect square $(2m^2)$.

Vidyanshu Mishra
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