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.
On
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.
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.