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Equation without a proof

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

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?

Note by William Isoroku
10 months, 1 week ago

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Let $\left \lfloor \frac{(2n \pm 1)^2}{2} \right \rfloor = v$ 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! $\large Q. E. D.$ · 10 months, 1 week ago