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Pythagorean triple is a triple of integers $(a, b, c)$ such that $a^2+b^2=c^2$. Is there any Pythagorean triple such that, not only $a^2+b^2$, but also $b^2+c^2$ is a square number? If not, how to prove it?

I tried to prove non-existence the following way: If true, it would mean that there is a pair of integers such that both sum and difference of their squares is a square number. Let's call these integers $a$ and $b$ and $a<b$. Then, there are integers $c$ and $d$ such that: \begin{align} &a^2+b^2=c^2 \\ &a^2-b^2=d^2 \end{align}

Multiplying those equations gives: \begin{equation} a^4=(cd)^2+b^4 \end{equation}

This is similar to Fermat's Last Theorem for $n=4$, but using it only shows that $cd$ can't be square number, not that there are no integer solutions.

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  • $\begingroup$ There are Pythagorean triples. $x^2+y^2=a^2+b^2=z^2$ And there are none. It still Euler proved. $\endgroup$ – individ Oct 6 '14 at 14:28
  • $\begingroup$ @individ could you please clarify your comment? $\endgroup$ – Danijel Oct 6 '14 at 14:42
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    $\begingroup$ @individ And how to prove that there are no solutions? $\endgroup$ – Danijel Oct 6 '14 at 14:44
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    $\begingroup$ Proof. $\endgroup$ – Lucian Oct 6 '14 at 15:40
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    $\begingroup$ See also: math.stackexchange.com/questions/1146460/… $\endgroup$ – Martin Sleziak Feb 15 '15 at 17:06
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These are triples with $x^2 + y^2 = 2 z^2,$ all positive and $\gcd(x,y) = 1.$

February 14, 2015   x^2 + y^2 = 2 z^2
           x           y           z
           1           1           1
           1           7           5
           1          41          29
           1         239         169
           7          17          13
           7          23          17
           7         103          73
           7         137          97
          17          31          25
          17          73          53
          17         193         137
          17         431         305
          23          47          37
          23          89          65
          23         289         205
          31          49          41
          31         151         109
          31         311         221
          41         113          85
          41         119          89
          47          79          65
          47         217         157
          47         497         353
          49          71          61
          49         257         185
          49         457         325
          71          97          85
          71         391         281
          73         161         125
          73         263         193
          79         119         101
          79         401         289
          89         191         149
          89         329         241
          97         127         113
         103         271         205
         103         313         233
         113         217         173
         113         463         337
         119         167         145
         119         233         185
         119         479         349
         127         161         145
         137         367         277
         137         409         305
         151         343         265
         161         199         181
         161         281         229
         167         223         197
         191         329         269
         193         497         377
         199         241         221
         217         353         293
         217         487         377
         223         287         257
         233         383         317
         241         287         265
         281         433         365
         287         337         313
         287         359         325
         337         391         365
         359         439         401
         391         449         421
jagy@phobeusjunior
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  • $\begingroup$ I'm going to find the notebook I did the scratch work in. But I'm pretty sure all those cases fail for the same reason, that $z^2 - x^2$ is not a perfect square. Edit: I just checked, all of those cases fail for that case. $\endgroup$ – Axoren Feb 14 '15 at 19:54
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For the $(a,b,c)$ and $(b,c,d)$ Pythagorean triples to work, we note:

  1. $a,b,c$ is primitive (otherwise reduce by the gcd)
  2. likewise $b,c,d$ is primitive
  3. Using the traditional solution for a triple: $2mn$ ; $m^2-n^2$ ; $m^2+n^2$, then $c$ is odd
  4. if $c$ is odd in the $(b,c,d)$ triple, then $b$ must be even = $2mn$
  5. Then we have ($2mn$, $m^2+n^2$, $d$ ) for the second triple
  6. Then $(2mn)^2$ + $(m^2+n^2)^2$ = $m^4 + 6m^2n^2 + n^4 = d^2$ which is a $\square$

But H.C. Pocklington of St John's College in 1913 proved that 6 for the coefficient of $x^2y^2$ in the general equation $x^4 + dx^2y^2+y^4 = z^2$ is impossible by prime moduli, in this case 6 != 7 mod 8 See H. C. Pocklington. "Some diophantine impossibilities" Proc. Camb. Phil. Soc, 17: pf 110 – 118, 1914.

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  • $\begingroup$ (typo?). I don't understand "6!=7 mod 8". $\endgroup$ – DanielWainfleet Mar 30 '17 at 22:03
  • $\begingroup$ @DanielWainfleet It means $6!=720=8\cdot90$ cannot be of the form $8n+7$. $\endgroup$ – Antonio DJC Jan 6 at 9:27

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