Does the equation, x^2 + y^3 = z^4 have solutions in prime numbers? Find at least one if yes, give a nonexistence proof otherwise.

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Does the equation, x^2 + y^3 = z^4 have solutions in prime numbers? Find at least one if yes, give a nonexistence proof otherwise.

via Math puzzles

Since we are working over prime numbers, precisely one of x,y,z is 2.

z=2 trivially gives no solutions.

If x=2, we need 4+y^3=z^4

y^3 = z^4-4 = (z^2-2)(z^2+2)

This limits the possibilities, as y^2 has to be z^2+2 and y has to be z^2-2, giving no solutions, or -y^2 has to be z^2-2 and -y has to be z^2+2, again giving no solutions.

If y=2, we need x^2+8=z^4

Again, this gives z^4-x^2=(z^2-x)(z^2+x)=8. Again, either z^2+x=4 and z^2-x=2, giving no solutions, or z^2-x=-4 and z^2+x=-2, again giving no solutions.

Hence no solutions in primes x,y,z.

Can you please explain why precisely one of x,y,z = 2? Thanks!

Check the parity of the equation...