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Recall that a perfect power has the form $x^m$ with $m,x\in\{2,3,\ldots\}$. Motivated by Fermat's result that the equation $x^4+y^4=z^2$ has no positive integer solution, here I ask the following question.

Question 1. Can $x^4+y^4+1$ with $x,y\in\mathbb N=\{0,1,2,\ldots\}$ be a perfect power?

Based on my computation, I conjecture that $x^4+y^4+1$ with $x,y\in\mathbb N$ can never be a perfect power.

Question 2. Can we find $x,y\in\mathbb N$ such that $x^4+y^4+1=\prod_{i=1}^kp_i^{a_i}$ for some $a_1,\ldots,a_k\in\{2,3,\ldots\}$ and distinct primes $p_1,\ldots,p_k$?

Via a computer I find no $x^4+y^4+1$ with $x,y\in\{0,1,\ldots,8000\}$ of the form $\prod_{i=1}^kp_i^{a_i}$ with $p_1,\ldots,p_k$ distinct primes and $a_1,\ldots,a_k\in\{2,3,\ldots\}$. Perhaps, Question 2 has a negative answer. Of course, a negative answer to Question 2 implies a negative answer to Question 1.

Your comments are welcome!

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    $\begingroup$ Numbers as in Question 2 are known as powerful numbers. Related question, which may suggest that the answer is positive but the least example is large. $\endgroup$
    – Wojowu
    May 15 at 14:52
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    $\begingroup$ It has solutions in rationals like $$(\frac{95800}{414560})^4+(\frac{217519}{414560})^4+1=(\frac{422481}{414560})^4.$$ $\endgroup$
    – Max Alekseyev
    May 15 at 16:31
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    $\begingroup$ Related: mathoverflow.net/questions/61794/the-diophantine-eq-x4-y4-1-z2 $\endgroup$
    – Thomas Browning
    May 15 at 17:59
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    $\begingroup$ Note that $x^4+y^4+1\equiv2{\rm\ or\ }3\bmod5$ (so is not a perfect square) unless $x$ and $y$ are both multiples of five. $\endgroup$
    – Gerry Myerson
    May 16 at 1:18
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    $\begingroup$ Please restrict to one question per post. $\endgroup$
    – GH from MO
    May 16 at 8:34
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To answer question 2: $$346^4+36788^4+1=1831575032204939793=3^3\cdot19^3\cdot179^2\cdot17569^2.$$

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  • $\begingroup$ Cool. How did you calculate this? $\endgroup$
    – Mitch
    May 15 at 18:32
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    $\begingroup$ I just ran through possibilities for $x$ and $y$, and then did trial division. Usually with trial division you stop when $p^2>n$, but if you're checking whether a number is powerful then you can stop when $p^5>n$, and then check whether $n$ is a perfect square or perfect cube. $\endgroup$
    – Thomas Browning
    May 15 at 18:45
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    $\begingroup$ One can use PARI/GP with a built-in function ispowerful(). $\endgroup$
    – Max Alekseyev
    May 15 at 18:50

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