This is as far I got:
y^2+108=x^3
y^2+100=x^3-8
y^2+100=(x-2)(x^2 +2x +4)
And now im a bit stuck
#Prove that for every integer y, y^2 + 108 is not equal to the cube of an integer
clear mason
amber sorrelBOT
pine zinc
This is as far I got:
y^2+108=x^3
y^2+100=x^3-8
y^2+100=(x-2)(x^2 +2x +4)
And no...
First you prove that x cannot be even (bc 5 is not a square mod 16). So, x and y are odd. Then x^2+2x+4=(x+1)^2+3 is odd, so there is a prime divisor p=4k+3 dividing y^2+100 which means that -1 is quadratic residue mod p which is a contradicion.
hidden lantern
*Theorem in Apostol book (p191) that says that such an equation (mordell equations k + x^3 = y^2) has no solution if k
has the form k=(4n−1)^3−4m^2, with m
and n integers such that no prime p≡−1(mod4) divides m.
idk, but this might give you some tools for similar problems
have in mind that most of this problems are solved via congruences
try to solve x^3 + 23 =y^2
using the theorem and without using it