How do i find all integers $y$ such that $y^3 = 3x^2+3x+7$, where $x$ is also an integer?

Question

I have tried to find all integers $y$ such that $$y^3 = 3x^2+3x+7$$, where $x$ is also an integer but i didn't succed only i guess that no integer $y$

$x$satisfied that equation so i would like to solve to find all integers

satisfiing it ?

Thank you for any help .

Answer 1

Try to take mod $3 \rightarrow y^3 \equiv 1 \pmod {3} \rightarrow y \equiv 1 \pmod {3} \implies y^3 \equiv 1 \pmod {9}$

Therefore $3x^2 + 3x + 7 \equiv 1 \pmod {9} \rightarrow 3x^2 + 3x \equiv 3 \pmod {9} \rightarrow x^2 + x \equiv 1 \pmod {3}$

This is impossible. If $x \equiv 1 \pmod {3}, x^2 + x \equiv 2 \pmod {3}$ If $x \equiv 2 \pmod{3}, x^2 + x \equiv 0 \pmod{3}$ and obviously if $x \equiv 0 \pmod {3}, x^2 + x \equiv 0 \pmod{3}$

Therefore no such x and y exist where $x$ and y are both integers