TITLE: Can a Pythagorean Triple ever have TWO identical exponents > 1?
QUESTION [5 upvotes]: For example, the triple
$(12, 16, 20)$
has these exponents:
$(12^1, 4^2, 20^1)$ or $(12^1, 2^4, 20^1)$
denoted by
$(1,2,1)$ or $(1,4,1)$

Here are more triples (ignore simple (1,1,1) cases):

I ran millions of triple tests looking for two identical exponents > 1, but found no matches.
Is there an existing proof or can someone prove it?
Thanks!

REPLY [4 votes]: We do know that all three exponents can assume different values greater than $1$.  Here a solution is constructed from the $(3,4,5)$ triple.
Multiply that triple by $9$ to get $(27,36,45)$.  This has one leg with exponent $2$ and the other leg with exponent $3$.  The hypoteneuse has no exponent greater than $1$, but now we take advantage of the fact that multiplying the $(27,36,45)$ triple by any sixth power preserves the exponents of $2$ and $3$ in the legs.  We choose $45^6$ as the multiplier and arrive at
$(6075^\color{blue}{3}, 546750^\color{blue}{2},45^\color{blue}{7}).$