The Diophantine equation x²-y³=7.

We will prove there are not integer solution to this equation.

Suppose that (x,y) is a possible solution of the Diophantine equation. If y is even, reducing modulo 8 we would have x²=7 mod 8, which is absurd because 7 does not fit among the squares modulo 8. It follows that y is odd. Then from the fact that:

x²-y³=7

we obtain:

x²+1=y³+8=y³+2³=(y+2)(y²-2y+4)

Now since y²-2y+4=(y-1)²+3= 4k+3 for some k there exists a prime divisor p of y²-2y+4 with p=3 mod 4 (if all the divisors of 4k+3 were of the form 4m+1 their product would still have the form 4n+1 which is false). It follows:

x²=-1 mod p

and therefore, by a known property of numbers:

-1=(-1)^((p-1)/2)

from which necessarily p=1 mod 4, which is a contradiction because we had said p=3 mod 4 .