# Are there any positive integers $a, b, c, d$ such that both $(a, b, c)$ and $(b, c, d)$ are Pythagorean triples?

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: $$a^4=(cd)^2+b^4$$

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.

• There are Pythagorean triples. $x^2+y^2=a^2+b^2=z^2$ And there are none. It still Euler proved. – individ Oct 6 '14 at 14:28
• @individ could you please clarify your comment? – Danijel Oct 6 '14 at 14:42
• @individ And how to prove that there are no solutions? – Danijel Oct 6 '14 at 14:44
• – Lucian Oct 6 '14 at 15:40
• – Martin Sleziak Feb 15 '15 at 17:06

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

• 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. – Axoren Feb 14 '15 at 19:54

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.

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