\[{ (2n\pm 1) }^{ 2 }+{ \left\lfloor \frac { { (2n\pm 1) }^{ 2 } }{ 2 } \right\rfloor }^{ 2 }={ \left\lceil \frac { { (2n\pm 1) }^{ 2 } }{ 2 } \right\rceil }^{ 2 }\]

I somehow found an equation that can generate Pythagorean triplets as shown above for any integral value of \(n\), regardless of positive or negative.

But I can't find a formal proof... So any approach or suggestions?

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## Comments

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TopNewestLet \[\left \lfloor \frac{(2n \pm 1)^2}{2} \right \rfloor = v\] Then, \[\left \lceil \frac{(2n \pm 1)^2}{2} \right \rceil= v+1\\ (2n \pm 1)^2= 2v+1\]

Hence, the conjectured equation is equivalent to \[(2v+1)+(v)^2=(v+1)^2\] Which is obviously true! \[\large Q. E. D. \]

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Look at the 2n+1 case. (2n+1)^2=4n^2+4n+1. As in the expression, we are dividing by 2. The 1/2 will be canceled when the floor is taken, so we will have (2n^2+2n)^2 for that part. The LHS will then become (2n+1)^2+(2n^2+2n)^2. Now we have the RHS. Applying a similar method we get 4n^2+4n+1 to round up with the ceiling, so now we have (2n^2+2n+1)^2. Now equate the LHS and RHS, yielding (2n+1)^2+(2n^2+2n)^2=(2n^2+2n+1)^2. Dy difference of square we know this is true because the (2n+1)^2 must equal (2n^2+2n+2n^2+2n+1). Hence proven. The other case simplifies in a nearly identical way. If you need more specifications, just ask. Nice formula.

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