Can $406014677132263504491682$ be the sum of two fourth powers?

Question

Can $406014677132263504491682$ be the sum of two fourth powers? It may be so. Can anyone use Wolfram Mathematica , SAGE , or some computer program, to check whether this number is the sum of two fourth powers ?

The complete factorization of this number is given by : $406014 677132 263504 491682 = 2 × 1459 × 6883 × 21529 × 938976 705857$

Answer 1

You may want to try some program like this in future for testing.

However, I cannot say anything myself about how the program handles precision, etc, with taking $4$th powers and roots of large numbers, but at least you can be sure it's possible if the program returns with a positive (as you can verify it yourself).

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https://js.do/

<script>

var a = 406014677132263504491682;

var b = Math.pow(a / 2.0, 0.25);

var c = -1;

for (var i = 1; i <= b; i++) {

    if (Math.pow(a - Math.pow(i, 4), 0.25) % 1 == 0) {c=i; break;}
}

document.write(c);

</script>

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Some quick testing with random examples such as $32$ and $280286069726155265499093303843106=12371235^4+129387197^4$ seem to work, though.

Answer 2

Hmmmm. $1459^1$. Any whole number where a $4n-1$ prime has an odd exponent can't be rendered as a sum of squares. Let alone fourth powers.