Find solutions

[MM]

Find all solutions to the equation [tex]2x^7+11y^7=43z^7[/tex] in natural numbers.

[Guest]

There are no solutions in the integers except [tex](x,y,z)=(0,0,0)[/tex].

Considering things mod 43 gives
[tex]2x^7+11y^7 \equiv 0[/tex] mod 43

Let us suppose [tex](x,y,z)[/tex] is a solution which is not all zeros and [tex]x,y,z[/tex] have no factors in common (other than 1). (If there is a common factor [tex]k[/tex] we can remove it by using [tex](x/k,y/k,z/k)[/tex] instead.)

Case 1: [tex]x[/tex] or [tex]y[/tex] is divisible by 43.
If [tex]x \equiv 0[/tex] mod 43 then we can conclude [tex]11y^7\equiv 0[/tex] and therefore [tex]y\equiv 0[/tex] mod 43 (since 43 is prime). Similarly if [tex]y \equiv 0[/tex] mod 43 then we can show [tex]x\equiv 0[/tex] mod 43. Consequently [tex]2x^7+11y^7[/tex] must be divisible by [tex]43^2[/tex], therefore [tex]43z^7[/tex] is also divisble by [tex]43^2[/tex], so [tex]z^7[/tex] is divisible by 43, as is [tex]z[/tex]. We have shown that [tex]x,y,z[/tex] have a common factor of 43 which is a contradiction.

Case 2: [tex]x[/tex] and [tex]y[/tex] are not divisible by 43.
[tex]2x^7+11y^7 \equiv 0[/tex] mod 43
rearranges to
[tex]2x^7 \equiv 32y^7[/tex] mod 43
since 43 is prime we can divide by anything that is not a multiple of 43, therefore
[tex](xy^{-1})^7 \equiv 16[/tex] mod 43
raising both sides to the power of 6 gives
[tex](xy^{-1})^{42} \equiv 2^{24}[/tex] mod 43
By Fermat's little theorem the left hand side is 1. So
[tex]1 \equiv 2^{24}[/tex] mod 43
Squaring both sides gives
[tex]1 \equiv 2^{42+6}[/tex] mod 43
[tex]2^{42}[/tex] is congruent to 1 by Fermat's little theorem. So the statement reduces to
[tex]1 \equiv 64[/tex] mod 43
which is clearly false.

Hope this helps,

R. Baber.