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I have to show that there are infinitely many natural numbers which are simultaneously a sum of two squares and a sum of two cubes but which are not a sum of two 6-th powers.

How to approach this question? Is there some simple condition that gives that a number cannot be the sum of two sixth powers?

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If we start with $n^3+5^3$ we have for sure a sum of two cubes. Since: $$ n^3+5^3 = (n+5)((n-3)^2+(n+16)) $$ if we set $n=m^2-16$ we have: $$ (m^2-16)^3 + 5^3 = (m^2-11)((m^2-19)^2+m^2) $$ and if we set $m=l+6$ we have: $$ (l^2+12 l +20)^3 + 5^3 = ((l+5)^2+2l)((l^2+12 l+17)^2+(l+6)^2) $$ and if $l$ is twice a square, the LHS is a sum of two squares, since the set of numbers that are the sum of two squares is a semigroup due to Lagrange's identity. By setting $l=2q^2$, we have: $$ (4q^4+24q^2+20)^3+5^3 = ((2q^2+5)^2+(2q)^2)((4q^4+24q^2+17)^2+(2q^2+6)^2)$$ but if we assume $q\equiv 0\pmod{7}$, the LHS is $\equiv 5\pmod{7}$, hence it cannot be the sum of two sixth powers, since $n^6\in\{0,1\}\pmod{7}$.


A more sophisticated, non constructive argument may be the following one. In the range $[1,N]$, with $N$ being a huge number, there are at least $\frac{1}{4} N^{2/3}$ integers that are the sum of two positive cubes. Among these numbers, at least $\frac{1}{12\log^3 N} N^{2/3}$ are the sum of two squares (we may apply a rough sieve with respect to the primes $\equiv 3\pmod{4}$). However, in the interval $[1,N]$ there are no more than $N^{1/3}$ numbers that are the sum of two sixth powers, hence infinitely many numbers are the sum of two squares and the sum of two cubes, but not the sum of two sixth powers.

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