# Prove, a^2+b^2=c^2,there exists only 1 case such that a,b,c are consecutive non negative integers(3,4,5) [closed]

I want to prove, $$a^2+b^2=c^2$$,there exists only 1 case such that a,b,c are consecutive non-negative integers(3,4,5).

## closed as off-topic by David Richerby, xskxzr, Yuval Filmus, Kyle Jones, EvilOct 8 at 21:09

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• I'm voting to close this question as off-topic because it is purely mathematical and has no computational content. It should be asked on Mathematics. – David Richerby Oct 8 at 7:36
• Hint: assume that the solutions are $n$, $n+1$ and $n+2$. – David Richerby Oct 8 at 7:36
• ... and note that the restriction to non-negative integers rules out the alternative solution $(a,b,c)=(-1,0,1)$ – gandalf61 Oct 8 at 8:16
• It is not a lemma. A lemma is a theorem that is not important itself, but is just a building block for the theorem you are really interested in. – gnasher729 Oct 8 at 10:27

$$n^2 + (n + 1)^2 = (n + 2)^2 \Rightarrow n^2 + n^2 + 2n + 1 = n^2 + 4n + 4 \Rightarrow 2n^2 2n + 1 = n^2 + 4n + 4 \Rightarrow n^2 - 2n - 3 = 0 \Rightarrow n = -1, 3$$