Proof by contradiction: 1. Suppose $x^3 + 3 =4y(y+1)$ has an integer solution
Then $x^3 + 3 =4y^2+4y$
Then $x^3 + 3 + 4 = 4y^2 + 4y + 4$
Then $x^3 + 7 = (2y + 2)^2$
Not sure how to simplify it further...
These are two odd numbers with no common factor.
Since their product is a cube, they must both be cubes.
Hint: $x^3+3=4y(y+1)$ implies $x^3=4y^2+4y-3=(2y+3)(2y-1)$, so $x^3$ is the product of two odd numbers that differ by $4$. Can two such numbers have any factors in common?