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I don't know how to find solutions to this problem other than trial and error. I appreciate all responses.

  1. Are there Pythagorean triples whose difference also yields a perfect square?

$(a,b,c)$ such that $(a^2+b^2=c^2) \land (b \gt a) \land$ $(b^2−a^2=d^2), a,b,c,d\in \Bbb N$

  1. Is there also a solution where $a^2$ or $b^2$ is a perfect squares of a prime, and also $c^2$ and $d^2$ are perfect squares of a prime? (three out of four values)

Thanks!

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    $\begingroup$ rewording may help you to focus the search, for the first question, you are looking for Pythagorean triples $(a,b,c)$ / ($a^2+b^2=c^2) \land (b \gt a)$ (ordered) and at the same time $b^2-a^2=d^2$, $a,b,c,d \in \Bbb N$ $\endgroup$ – iadvd Sep 15 '15 at 1:28
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    $\begingroup$ done, now the text is complete I think, good luck! Some months ago I did also a question regarding squares and primes, it is not exactly related with your question, but if you are interested in this kind of topics, you can get insights about where to focus in your research. math.stackexchange.com/questions/1218279/… $\endgroup$ – iadvd Sep 15 '15 at 1:51
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    $\begingroup$ I'm looking at the list of Pythagorean triples and it appears that they must contain at least one even number, so all four values cannot have perfect squares whose factors are prime. $\endgroup$ – Tony Sep 15 '15 at 2:07
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    $\begingroup$ It's been known since antiquity at all Pythagorean triples include either one or three even numbers, $\endgroup$ – MY USER NAME IS A LIE Sep 15 '15 at 2:08
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    $\begingroup$ Related: math.stackexchange.com/questions/43519/… $\endgroup$ – Jack D'Aurizio Sep 15 '15 at 2:20
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Consider that $(d^2,d^2+a^2,d^2+2a^2)=(d^2,b^2,c^2)$.

The answer to your first question is then in the first lines of this Wikipedia entry.

Fermat's right triangle theorem: If three square numbers form an arithmetic progression, then the gap between consecutive numbers in the progression cannot itself be square.

Fermat's proof (through infinite descent) can be found on the same page.

Equivalent results are that $1$ is not a congruent number, or that the only rational points on the elliptic curve $$y^2 = x(x-1)(x+1)$$ are the trivial ones. Some references:

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  • $\begingroup$ What about triples that are not arithmetic progressions? $\endgroup$ – Tony Sep 15 '15 at 14:35
  • $\begingroup$ @lurker: what does your question mean? Generic triples of squares? Yes, they exist. There are also triples of squares in AP, but with common difference that is not a square, like $1,25,49$, for instance. $\endgroup$ – Jack D'Aurizio Sep 15 '15 at 14:42
  • $\begingroup$ The theorem states that if they form an arithmetic progression then the gap cannot be square, so I assume this applies to my question in that all Pythagorean triples that are also arithmetic progression cannot have difference that is a perfect square. I wanted to know if this also applies in all instances where it is not an arithmetic progression. $\endgroup$ – Tony Sep 15 '15 at 15:28
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    $\begingroup$ @lurker: so, if I understood correctly, you are asking if there is an infinite number of triples $(a,b,c)$ such that $a^2+b^2$ is a square and $b^2+c^2$ is a square. In such a case, the answer is clearly affirmative, by considering the general form of a primitive pythagorean triple. If we take $a=(4u^2-v^2),b=(4uv),c=2(u^2-v^2)$, then $a^2+b^2=(4u^2+v^2)^2$ and $b^2+c^2=(2u^2-2v^2)^2$ as wanted. $\endgroup$ – Jack D'Aurizio Sep 15 '15 at 15:57
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    $\begingroup$ @lurker: if both $a^2+b^2$ and $a^2-b^2$ are squares, then $a^2-b^2,a^2,a^2+b^2$ is a 3AP of squares having a square as common difference. That is discussed above. $\endgroup$ – Jack D'Aurizio Sep 15 '15 at 20:26

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