$x^4+y^4$ powerful for relatively prime $x,y$

Question

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.

Answer 1

Disclaimer: This is not a complete solution but some partial results that may be helpful for a complete solution. Also note that this is my first contribution to Math Overflow so I hope I follow all guidelines and I would appreciate tolerance and feedback if I did something wrong.

I couldn't find any further progress on this problem on the internet (other than this thread). I believe I can reduce the search space to about $\approx 3.5*10^8$ cases.

Any powerful number can be written in the form $mz^2$ with $m|z$, so we are trying to solve $x^4 + y^4 = mz^2$. Since $m|z, x^4 + y^4 > m^3$, we have to understand what happens for all $m < N^{1/3} ~ 1.5 * 10^{12}$. Now $m$ must obey the following conditions (this is how I reduce the search space):

• $m$ must be squarefree.
• $m$ must be odd. Since $gcd(x,y) = 1, x^4 + y^4$ is either $1$ or $2 \mod 16$, but a powerful number is clearly not $2 \mod 16$.
• every prime factor $p$ of $m$ must be $1 \mod 8$, since if $p$ divides $x^4 + y^4 $but not $x$ or $y$, then $(x/y)^4 \equiv -1 \mod p$, so $p \equiv 1 \mod 8$.
• $2m$ must be a congruent number. Indeed, suppose $(x,y,z)$ is a solution to $x^4 + y^4 = mz^2$. Let $u = -4x^2y^2/z^2$, and let $v = 4xy(x^4-y^4)/z^3$. Then $v^2 = u^3-4m^2u$, so the congruent number curve $E(2m)$ is a principal homogenous space for the quartic $x^4 + y^4 = mz^2$.

Therfore we need to check curves with those odd $m$ such that each prime factor of $m$ is congruent to $1 \mod 8$, such that $2m$ is a congruent number which is $2 \mod 16$.

It is also known that there are no relatively prime solutions to $x^4 + y^4 = m^3$ (By Proposition 14.6.6 of Cohen's Number Theory: Volume II: Analytic and Modern Tools).

The next possibility, is $x^4 + y^4 = 17^2 m^3$. This means, we only need to check $m$ up to $(N/17^2)^{1/3} \approx 2.2 * 10^{11}$, i.e., $2m$ up to $\approx 4.4* 10^{11}$. By Table 3 of http://homepages.warwick.ac.uk/~masfaw/congruent.pdf (which counts the number of congruent numbers which are $2 \mod 16$ up to $10^{12}$), there are $59536672 + 62455317 + 66579936 + 73445274 + 81759844 + 12294626 + 2110645 = 358182314$ curves to check.

Answer 2

If coprime integers $x, y$ satisfy $x^4 + y^4 = A C^2$ with $C = A B$ then no prime 3 mod 4 divides the LHS, and that implies $A = a^2 + b^2$ and $B = c^2 + d^2$.

Also, if $x, y$ are coprime then $x^4 + y^4$ is odd or twice odd, so that $A$ and $B$ are also odd, and thus $x^4 + y^4$ is odd.

Furthermore, we can assume that $A$ is squarefree because any squared factor can be absorbed in $C^2$ and then in $B$, and $A$ being squarefree implies that $(a, b) = 1$.

One's first thought is simply to assume that $x^2 + i y^2 = (a + i b)^3 (c + i d)^2$ and search for integer solutions $x, y$ by looping through ranges of $a, b, c, d$.

Another possible search strategy is to consider the equation in the form

$x^4 + y^4 = (a (a^2 + b^2) B)^2 + (b (a^2 + b^2) B)^2$.

Given that $x, y$ are coprime, there must be integers $p, q, r, s$ with $(p, s) = (q, r) = 1$ and pairs $p, s$, $q, r$ of opposite parity, such that:

$x^2, y^2, a(a^2 + b^2) B, b(a^2 + b^2)B = p q - r s, p r + q s, p q + r s, p r - q s$

So one could search over sets of $p, q, r, s$ where for each set the procedure would be as follows:

1. Determine $P, Q = p q + r s, p r - q s$ and $R, S = p q - r s, p r + q s$
2. Perform the following in parallel, and abort step in the event any test applies:
   • Denoting $T = (P, Q)$ if $(P/T)^2 + (Q/T)^2$ does not divide $T$
   • For small(ish) primes, e.g. $2, 3, 5$, if $2^{2\alpha+1}$ || R or S etc
   • odd part of |R| or |S| not = 1 mod 8, non-3-divisible part of |R| or |S| not = 1 mod 3, etc, i.e. quadratic residue checks
   • |R| or |S| not a perfect square
3. Log solution we have found!