0
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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...

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  • $\begingroup$ How about considering your options $\mod 4$? $\endgroup$ – Laars Helenius Oct 14 '15 at 13:18
  • $\begingroup$ Your last step is wrong: $(2y+2)^2 = 4y^2+8y+4$ $\endgroup$ – gammatester Oct 14 '15 at 13:29
5
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$x^3=4y^2+4y-3=(2y+3)(2y-1)$
These are two odd numbers with no common factor.
Since their product is a cube, they must both be cubes.

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  • $\begingroup$ how is that a contradiction? You can have odd cubes that don't have common factors like 3^3 and 5^3? $\endgroup$ – shoestringfries Oct 14 '15 at 14:13
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    $\begingroup$ But they don't differ by 4. The difference between 1,8,27,64 grows quickly. $\endgroup$ – Empy2 Oct 14 '15 at 14:19
3
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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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