Can two the sum of two primitive perfect squares be equal? $a^2+b^2 = c^2 + d^2$?

Question

Given a primitive perfect square, $n^2=a^2+b^2$ where $gcd(a, b) = 1$ and
$m^2=c^2+d^2$ where $gcd(c,d)=1$
Can $n=m$?

a, b, c, d, n, and m are positive integers. And I forgot to mention:
EDIT 1: Two of (n, a, b) are divisible by 3. One of (n, a, b) is divisible by 9. And the same for m, c, and d: Two of (m, c, d) are divisible by 3. One of (m, c, d) is divisible by 9.

I have not found a situation among the first few hundred lowest perfect squares where $n=m$ and I would like to make sure this is true.

I know that any square can be represented by a sum of odd numbers, $$n^2=\sum_{k=1}^n (2k-1)$$

I find it interesting that none are equal.

Since $n=m$ is exactly equivalent to $n^2 = m^2$ when $n$ and $m$ are positive, why do you ask about "$n=m$ or $n^2=m^2$" as if they were two independent questions? — Erick Wong, Jul 16 '17 at 19:03
@Erick - sloppy typing. And no, I've not studied the subject — Stoddard, Jul 16 '17 at 19:27

score 3 · Accepted Answer · answered Jul 16 '17 at 18:11

3

You mean, like can something like $63^2+16^2$ equal something like $33^2+56^2$ and both be squares?

I'll have to think about that!

answered Jul 16 '17 at 18:11

Angina Seng

158,341

Because $65=(1^2+2^2)(4^2+3^2)$ There are infinitely many these solutions. – Michael Rozenberg Jul 16 '17 at 18:38
I forgot to include the mod3, mod9 constraints. – Stoddard Jul 16 '17 at 19:22
2

@Stoddard I am confused by the constraints you added: if any two of $(n,a,b)$ are divisible by $3$ then so is the third, which prevents these from being primitive. More generally, it is easy to show that if $n^2$ is a primitive sum of two squares then it is never divisible by $3$. – Erick Wong Jul 16 '17 at 19:30

score 0 · Answer 2 · answered Jul 16 '17 at 20:17

I cannot forget that scene in the movie "Love in the quadratic reciprocity era" where

$$ 3145^2 = 2263^2+2184^2 = 1463^2+2784^2 = 336^2+3127^2 $$ is mentioned.

Just joking. Such identities can be easily derived from Lagrange's identity $$ (a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2 $$ and the fact that every prime $p\equiv 1\pmod{4}$ can be represented in a essentially unique way as $u^2+v^2$, since $\mathbb{Z}[i]$ is a Euclidean domain, hence a UFD. It follows that by taking some square-free $n$ given by the product of many primes $\equiv 1\pmod{4}$, $n^2$ can be represented as $u^2+v^2$ (with $\gcd(u,v)=1$) in many ways. Lord Shark's choice is $n=5\cdot 13$ and my choice above is $n=5\cdot 17\cdot 37$. To deal with the $\!\!\pmod{9}$ constraints is simple.

lol. That may have been the taxi cab scene that I missed. https://en.wikipedia.org/wiki/Generalized_taxicab_number — Stoddard, Jul 16 '17 at 20:37

Can two the sum of two primitive perfect squares be equal? $a^2+b^2 = c^2 + d^2$?

2 Answers2