# Question about Pythagorean Triples with the same $c$ value

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 $$a_{0}-a_{1}=a_{3}$$

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