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Suppose that $(a_{0},b_{0},c)$ , $(a_{1},b_{1},c)$ , $(a_{2},b_{2},c)$ are Pythagorean triples with the same $c$ value and completely different $a$'s and $b$'s. I'm trying to prove that $a_{0}+a_{1}=a_{2}$ isn't possible.

For example if c = 65, we can factor $c^{2}$ into primes $5\cdot5\cdot13\cdot13$ which can be factored further into gaussian primes $(2+i)(2-i)(2+i)(2-i)(3+2i)(3-2i)(3+2i)(3-2i)$. If you multiply them in 2 groups with the same size you will be left with 2 conjugate pairs, and by the formula $(a+bi)(a-bi)=a^{2}+b^{2}$ we can multiply them and get a sum of squares. For example, $[(2+i)(2+i)(3+2i)(3-2i)]\times[(2-i)(2-i)(3+2i)(3-2i)] = [39+52i]\times[39-52i]=39^{2}+52^{2}$ which does equal $65^{2}$.

If you do every combination you will get $39^{2}+52^{2}=65^{2}$, $33^{2}+56^{2}=65^{2}$, $25^{2}+60^{2}=65^{2}$ and $16^{2}+63^{2}=65^{2}$ , and if you try to match up the values and see if they correspond to any other, they don't.

Using this knowledge I tried brute-forcing solutions up to $c\approx2.2\times10^{6}$ and got nothing. I suspect this is impossible but have no clue how to prove it. Do you have any answer or what do you suggest I do next?

PS: In case a numerical solution is found, I would also like your consideration on the full problem, that has an extra triple $(a_{3},b_{3},c)$ and the extra condition $a_{0}-a_{1}=a_{3}$

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How about this:

$425^2+1020^2 = 576^2+943^2=1001^2+468^2=1105^2$

and $425+576=1001$

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  • $\begingroup$ I just realized I messed up the question, probably going to add a new post with the correct one, but thank you anyway. I made a severe mistake when writing the problem, but this one is still interesting nonetheless $\endgroup$ Jan 7 at 11:20

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