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I am looking for two integers, a and b , which when raised to the fourth power and added together equal either $115723683465520156810$, or $115712970354183898410$ , or $115734396970441439210$.

Thus $(a^4) + (b^4) = 115723683465520156810$ or $(a^4) + (b^4) = 115712970354183898410$ or $(a^4) + (b^4) = 115734396970441439210 $

Can anyone give me a software program to check whether an integer is the sum of two fourth powers?

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  • $\begingroup$ Your three numbers are divisible by $7$, $3$, and $23$, respectively, but not by the squares of those primes, each of which is congruent to $3$ mod $4$. They are consequently not expressible as a sum of two squares, much less two fourth powers. $\endgroup$ – Barry Cipra Mar 6 '17 at 0:26
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None of those equations are possible.

Suppose $a^4 + b^4 = n$ where $n$ is one of the three numbers

$$115723683465520156810,\;\;115712970354183898410,\;\;115734396970441439210$$

If either of $a,b$ is a multiple of $5$, then, since $n$ is a multiple of $5$, it follows that $a,b$ must both be multiples of $5$. But then $a^4+b^4$ would be a multiple of $5^4$, contradiction, since $n$ is a not a multiple of $25$. Thus, neither of $a,b$ is a multiple of $5$. But then \begin{align*} &a^4 + b^4 = n\\[6pt] \implies\; &a^4 + b^4 \equiv 0 \pmod{5}\qquad\text{[since $n$ is a multiple of $5$]}\\[6pt] \implies\; &1 + 1 \equiv 0 \pmod{5}\qquad\;\;\;\text{[by Fermat's little Theorem]}\\[6pt] \implies\; &2 \equiv 0 \pmod{5}\\[6pt] \end{align*}

contradiction.

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