Odd Pythagorian triplets [closed]

How many Pythagorean triplets $$\{a,b,c\}$$ exist, where $$a,b,c$$ are all odd?

As far as I know there are no such triplets. $$\{3, 4, 5\}; \{5,12,13\} ; \{7,24,25\}$$ and its multiples are examples.

Is there any explanation on why all the integers forming a triplet are not all odd or all even?

closed as off-topic by Namaste, Holo, Cesareo, Saad, DidDec 25 '18 at 8:45

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• Think about sums and squares of odds and evens. – Randall Dec 22 '18 at 15:24
• Yeah..if we consider sum o square of odds they wont be perfect squares( not integers) to valid for pythogorian triplet but then is there any way where we can proove this one with formulas are something instead of logic – Fathima Jasmine Dec 22 '18 at 15:31
• All even: 6, 8, 10. – ypercubeᵀᴹ Dec 22 '18 at 15:55
• Since $B=2mn$ and one of $m,n$ is even, $B$ is always a multiple of $4$ so you cannot find a triple with all odd digits. – poetasis Jun 15 at 19:18

Suppose the numbers are $$a,b,c$$. Suppose $$a^2+b^2=c^2$$. The order doesn't matter anyway. Since $$a$$ and $$b$$ are odd, $$a^2+b^2$$ will be even. But, $$c^2$$ is odd. Hence, no such triplet can exist.

$$a^2b^2(a^2+b^2)$$ mod $$2$$
$$a^2b^2(a^2+b^2+2ab)$$ mod $$2$$
$$a^2b^2(a+b)^2$$ mod $$2$$
If both $$a$$ and $$b$$ are such that $$a \equiv 1 \text{ mod } 2$$ and $$b \equiv 1 \text{ mod } 2$$ then $$a+b \equiv 0 \text{ mod } 2$$ hence abc is divisible by 2.
Use that $$a^2+b^2=c^2\iff(\lambda^2-\mu^2)^2+(2\lambda\mu)^2=(\lambda^2+\mu^2)^2$$