# Can we find primitive Pythagoras triplet (in integers), with two sides being as powerful numbers?

Actually this seemingly elementary question had very much surprised me, I though it is quite easy to find such triples described in question, but strangely after trying so many times, couldn't find any, but the hope is there in relatively large solutions, or may be a notable work is there about this issue where I don't encounter yet.

A powerful number by definition is an integer which has all exponents in its prime factors greater than one.

The reference of this question was originally in Quora.

• The Quora question is more general than the one you have asked. In fact you don't seem to have asked a question in the body. In the subject, do you mean for any two numbers to be powerful, or only the 'first' two ($x^m$ and $y^n$, in the notation of the Quora question)? – LSpice Jun 12 '16 at 22:11
• @LSpice I think the question was slightly different at Quora in the above link, where the confusion comes from the definition of powerful number without distinct deference from being a square or a cube or an integer raised for another integer exponent, yes that was stating clearly $x^{2n} + y^{2m} = z^k$, has no integer solution if $x, y, z$ are coprime integers, where $n, m, k$ are positive integers > 1, of course the notable case is when $k = 2$, but then he made the confusion describing them as powerful numbers, I don't see why this might be true, there may be a counter example for this ! – Sophyan Gharz Jun 14 '16 at 3:45
• I have added this question in relation to this one also here. – Sophyan Gharz Jun 23 '16 at 0:10

Yes, this follows from Elkies' question here: $x^4+y^4$ powerful for relatively prime $x,y$
Let $u=427511122^2,v= 1322049209^2$.
In the primitive triple $(u^2-v^2,2uv,u^2+v^2)$ the powerful sides are $2uv$ and $u^2+v^2$.
Almost surely there are infinitely many solutions via elliptic curves like $x^4+1=k^3y^2$ or intersection of quadrics.
• @SophyanGharz How Is not $2uv$ powerfull? $u,v$ are squares and $u$ is even? Did you try in on a computer? – joro Jun 13 '16 at 10:56
• Yea, I was reediting my comment , but was too late, I noticed that you are taking $u^2, v^2$, which is true as powerful number for $2u^2v^2$, but untrue for $u^4 - v^4$, then we must check $u^4 + v^4 – Sophyan Gharz Jun 13 '16 at 11:05 • @SophyanGharz This is what I wrote in the answer. Check$u^2+v^2\$ for u,v as I gave them, they are squares. Factoring is fast in this case). – joro Jun 13 '16 at 11:09