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

Proof by contradiction: 1. Suppose $x^3 + 3 =4y(y+1)$ has an integer solution

1. Then $x^3 + 3 =4y^2+4y$

2. Then $x^3 + 3 + 4 = 4y^2 + 4y + 4$

3. Then $x^3 + 7 = (2y + 2)^2$

Not sure how to simplify it further...

• How about considering your options $\mod 4$? – Laars Helenius Oct 14 '15 at 13:18
• Your last step is wrong: $(2y+2)^2 = 4y^2+8y+4$ – gammatester Oct 14 '15 at 13:29

$x^3=4y^2+4y-3=(2y+3)(2y-1)$
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?