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(From a math competition) Question: Find all integers $a,b,c$ that satisfy:

$$a^3 - 3a^2b - 3c+2b^2 = c^3 -3ab^2 + 3c^2 +1 $$

What I have tried/attempted basically I've been looking for expansions such as $(a+b)^3$ etc. and I could find one $(c+1)^3$

so

$$a^3 - 3a^2b - 3c+2b^2 = c^3 -3ab^2 + 3c^2 +1 $$

$$ \Leftrightarrow a^3 -3a^2b+3ab^2 +2b^2 = c^3+3c^2+3c+1 $$

$$ \Leftrightarrow a^3 -3a^2b+3ab^2 +2b^2 =(c+1)^3$$

But other than that I'm not sure how to continue. . . .

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bigfocalchord
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  • By subtracting $b^3$ from both sides we have $(b+2)b^2=(c+1)^3-(a-b)^3$ – Emre May 21 '16 at 12:08
  • I calculated $m^3-n^3=(b+2)b^2$ for $-1000\le m,n,b\le 1000$ and there are a lot of irregular-looking solutions. Are you sure about the question? – Emre May 21 '16 at 13:38
  • @e.girgin maybe the OP misread the question. It might be meant for Fermats last theorem – N.S.JOHN May 21 '16 at 13:41

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(Not yet allowed to comment so this has to be provided as an answer)

Continuing - but not necessarily solving…

$(a-b)^3+b^3+2b^2=(c+1)^3$

With $b=0$ this gives an infinite set of answers with $a=c+1$.

For the rest, rewriting as

$(c+1)^3-(a-b)^3={b^2}(b+2)$ (or let's say $x^3-y^3={b^2}(b+2)$)

gives a series of "difference of two cubes" problems. My guess is that these can be taken in stages and handled with modular arguments. But I have to admit that I haven't actually tried.

Martin Kochanski
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