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

Question

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...

How about considering your options $\mod 4$? – Laars Helenius Oct 14 '15 at 13:18 — 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 — gammatester, Oct 14 '15 at 13:29

score 5 · Accepted Answer · answered Oct 14 '15 at 13:29

5

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

answered Oct 14 '15 at 13:29

Empy2

52,372

how is that a contradiction? You can have odd cubes that don't have common factors like 3^3 and 5^3? – shoestringfries Oct 14 '15 at 14:13
1

But they don't differ by 4. The difference between 1,8,27,64 grows quickly. – Empy2 Oct 14 '15 at 14:19
See here for a proof that they must both be cubes. – Bill Dubuque Feb 02 '21 at 09:02

score 3 · Answer 2 · answered Oct 14 '15 at 13:29

3

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?

answered Oct 14 '15 at 13:29

Barry Cipra

81,321

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

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