In Honour of Helmut Hasse (2)

All the integers in the list may be written as $2002 + 4n$ so we are asked to solve the diophantine equation $x^7 + 4y^7 = 2002 + 4n$ . First, rewrite the equation as $x^7 - 2002 = 4(n - y^7)$ . If a solution existed then that solution would be a solution of all the corresponding congruences, so consider the congruence $\mod 4$ , $x^7 - 2002 \equiv 0 \mod 4$ . We can simplify this to $x^7 \equiv 2 \mod 4$ . If $\gcd(x,4) > 1$ then $x^7 \equiv 0 \mod 4$ so if a solution exists then $\gcd(x,4) = 1$ and thus we may apply the Euler-Fermat theorem with $\phi(4) = 2$ to reduce the congruence to $x \equiv 2 \mod 4$ . But this is a contradiction, because $\gcd(4,4k + 2) > 1$ . Therefore, no solutions exist.