Prove $x^3 + 3 =4y(y+1)$ has no integer solutions

elementary-number-theory algebra-precalculus

Solution 1:

$x^3=4y^2+4y-3=(2y+3)(2y-1)$
These are two odd numbers with no common factor.
Since their product is a cube, they must both be cubes.

Solution 2:

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?

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