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Is 38012 460109 621768 889218 the sum of two fourth powers? I suspect that this number is the sum of two fourth powers. Can anyone use Wolfram Mathematica or SAGE to check whether this number is the sum of two fourth powers ?

The complete factorization of this number is given by : 38012 460109 621768 889218 = 2 × 41 × 1217 × 21529 × 27617 × 640 650529 Notice that all the factors are of the form 16n+1 or 16n+9 . Hence this number may be the sum of two fourth powers.

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  • $\begingroup$ Just write a computer program that checks all the numbers $n$ from 1 to 441551, and check if $(38012460109621768889218-n^4)^{1/4}$ is an integer. $\endgroup$ – Jonathan Davidson Jul 8 '17 at 21:17
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    $\begingroup$ Observation: $x,y$ must both be odd if $x^4+y^4$ equals this number. The odd squares modulo $10$ are congruent to $1,5,9$. Upon squaring each of these, we get $1,5$. Therefore the number cannot be such a sum. $\endgroup$ – abiessu Jul 8 '17 at 21:17
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    $\begingroup$ JUst asked wolfram alpha which says there are no integer solutions $\endgroup$ – Bram28 Jul 8 '17 at 21:18
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The answers to the questions asked are "No" and "Yes" respectively.

By inspection of the factorization, the number is the double of an odd number. Therefore, we cannot have $(2x)^4+(2y)^4$ or $(2x)^4+(2y+1)^4$, as these both produce the wrong number of factors of $2$. Therefore the form we must have is $(2x+1)^4+(2y+1)^4$. We can rewrite this as $(10x+a)^4+(10y+b)^4$ for $a,b\in\{1,3,5,7,9\}$. From this we get that $a^2,b^2\in\{1,5,9\}\mod 10$ and further that $a^4,b^4\in\{1,5\}\mod 10$. Using a simplification of the binomial theorem, we get $(10x+a)^4+(10y+b)^4=10q+a^4+10r+b^4$, and since $a^4,b^4\in\{1,5\}\mod 10$, we get $a^4+b^4\in\{0,2,6\}\mod 10$, and so the specified number cannot be the sum of two fourth powers.

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  • $\begingroup$ You are right @abiessu. Great answer. Thanks for your analysis. Well Done !! $\endgroup$ – Derek Jul 8 '17 at 21:41
  • $\begingroup$ (+1) Simple and straight to the point, excellent answer. $\endgroup$ – Jack D'Aurizio Jul 8 '17 at 22:33
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The fourth power of any integer ends in $0,1,5$,or $6$, so you can't add two of these to get your number which ends in $8$.

Also, here is the relevant Wolfram Alpha query which confirms there are no integer solutions.

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