I asked this question on the NMBRTHRY mailing list on 17 February 2014, but it remains unsolved as far as I know.

Recall that a "powerful number" is a positive integer whose prime factorizations $m = \prod_i p_i^{e_i}$ has each exponent $e_i \geq 2$. (Equivalently $-$ though generally of little use $-$ a positive integer is powerful if and only if it can be written as $m^2 n^3$ for some integers $m,n$.) Pondering this Mathoverflow question led me to ask:

What's the smallest powerful number that can be written as $x^4+y^4$ with $\gcd(x,y) = 1$?

In particular, is it $3088257489493360278725196965477359217 = 427511122^4 + 1322049209^4$?

The gcd condition is needed for the usual reason: if $x^4+y^4$ is powerful then so is $(cx)^4+(cy)^4$, but the converse fails, and indeed any number $m$ can be made powerful by multiplying it by some $c^4$ (say $c=m$ itself); we are not interested in examples such as $17^4 + 34^4 = 17^5$.

There are only about twice as many powerful numbers $m \leq x$ as there are squares [the actual ratio is $A = \zeta(\frac32)/\zeta(3) = 2.17325+$, and if I did this right then the count is given more precisely by $A x^{1/2} - B x^{1/3} + o(x^{1/6})$, where $B = -\zeta(\frac23)/\zeta(2) = 1.48795+$, and the $o(x^{1/6})$ is actually $o(x^{1/12+\epsilon})$ under the Riemann Hypothesis; but this is all tangential to the question at hand].

Thus, as with squares, we expect only finitely many examples of
coprime $x,y$ for which $x^5+y^5$ is powerful, but *do* expect
$x^4+y^4$ to be powerful for for an infinite though sparse set of
coprime pairs $(x,y)$. True, Fermat showed that there are no solutions of
$x^4 + y^4 = z^2$; but there are integers $m$ for which the elliptic curve
$x^4 + y^4 = mz^2$ does have infinitely many rational points, and indeed
we can use such curves to find powerful $x^4 + y^4$: compute solutions of
$x^4 + y^4 = mz^2$ until finding one for which $z$ is divisible by
each prime factor of $m$. For example, taking $m=17$ eventually yields
$$
427511122^4 + 1322049209^4 = 17 \cdot 426218494746902449^2
= 17^3 \, 73993169^2 \, 338837713^2.
$$
This is the smallest example I found, but this method needn't
find solutions of "$x^4 + y^4 = $ powerful" in order of
increasing size, and I don't see how to organize an exhaustive search
that could provably find the smallest example if it is not much smaller
than the solution above (for which $x^4 + y^4 \doteq 3 \cdot 10^{36}$).
For what it's worth, Google does not recognize it.

By the way, it's much easier to search for powerful values of $x^4 - y^4$
(again with $\gcd(x,y)=1$), because $x^4-y^4$ factors, and each of the factors
$x+y$, $x-y$ must be powerful except possibly for a stray power of $2$.
This means that trying all $(x,y)$ with $x+y \leq H$ takes time
proportional to $H$. For instance, it took just over 6 hours of **gp**
computation to find that
$$
10113607^4 - 4319999^4 = 6 \cdot 41056761311940^2
= 2^5 \, 3^3 \, 5^2 \, 11^2 \, 23^2 \, 37^2 \, 47^2 \, 313^2 \, 4969^2
$$
is the only example with $x+y \leq 10^8$, even though
$x^4 - y^4$ is still quite large (just over $10^{28}$).
This example *is* known to Google, but only as a solution of $x^4-y^4=6z^2$,
with nothing about $6z^2$ being powerful, let alone about its being
the first such example.

gptocountto $10^{16}$, let alonefactorevery number of at most $16$ digits! Much better to try all coprime $(x,y)$ of opposite parity with $x<y$ and $x^4 + y^4 < 10^{16}$; that's only 40 million or so factorizations, which take a few hours to try (and as expected find nothing). Still it's hopeless to reach $3 \cdot 10^{36}$ this way... Now that it's a couple of months since I posted this question, I should post a partial answer evaluating different strategies, the best of which might make the computation barely feasible. $\endgroup$ – Noam D. Elkies Mar 11 '15 at 23:4510more comments