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# Will someone give me a LOGIC for this??

All of you may have observe that: 3^2+4^2=5^2 (if not then observe it now)

But instead, if you want to find two numbers such that sum of their squares is equals to square of any specific number then it would be a hard task to find them. But I have found myself a general way to find those numbers.

Let T be the number (say 13) for which you want two numbers m,n such that sum of their squares is equals to T^2 (in our example, 13^2=169). Now divide T by 5 and let the answer be a (13/5=2.6). Now:

m=T-a (m=13-2.6=10.4)

n=T-2a (n=13-2x2.6=7.8)

Here we got our numbers. Now check that (m^2+n^2) will be equals to T^2.

Although it worked for every number I have tried yet but I can't find an explanation for that why one have to divide every number with 5. So please comment any logic you can get for this and also your views about should I get an medal or any award for this discovery.

Note by Vaibhav Jain
2 years, 4 months ago

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You're basically exploiting the fact that $$(3k)^2 + (4k)^2 = (5k)^2$$. (This is your identity multiplied by $$k^2$$.)

From your formula, $$m = T - a = T - \dfrac{T}{5} = \dfrac{4T}{5}$$ and $$n = T - 2a = T - \dfrac{2T}{5} = \dfrac{3T}{5}$$

and therefore $$m^2 + n^2 = \dfrac{4T}{5}^2 + \dfrac{3T}{5}^2 = \dfrac{5T}{5}^2 = T^2$$.

You can do this using other Pythagorean triplets as well, for example, for some number $$S$$, let $$b = \dfrac{S}{13}$$. and defining $$k = S - b$$ and $$l = S - 8b$$, we can see that $$k^2 + l^2 = S^2$$ · 2 years, 4 months ago