I have a question for which I was not able to find an answer online. I was wondering how many Pythagorean Triplets we have found till now which consists of three consecutive integers like $(3,4,5)$.
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5There are no more. Try solving $x^2 + (x+1)^2 = (x+2)^2$. – player3236 Dec 08 '20 at 04:35
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1Similar but different question – J. W. Tanner Dec 08 '20 at 04:41
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@player3236 There are two solutions to the equation you showed in your comment. These are $(-1,0,1)$ and $(3,4,5)$ as I showed in my answer below. Granted one is trivial. – poetasis Dec 10 '20 at 02:24
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For slightly simpler numbers, let $x$ be the middle number then you have to solve $$ \begin{split} (x-1)^2 + x^2 &= (x+1)^2 \\ x^2 &= (x+1)^2 - (x-1)^2 = 2x\cdot2 \end{split} $$ which yields $x=0$ and $x=4$. Can you finish?
gt6989b
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$$x^2 +(x+1)^2=(x+2)^2\\\implies x^2 +(x+1)^2-(x+2)^2=0\\\implies x^2 - 2 x - 3 = 0$$
Using the quadratic equation, we find $$x\in\{3,-1\}\implies \text{the set of triples}\quad S=\{(3,4,5),(-1,0,1)\}$$ There are no other solutions.
J. W. Tanner
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poetasis
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