Let $a$, $b$, and $c$ be integers satisfying $a^2 + b^2 = c^2$. Prove: $abc$ must be even. I'm pretty sure that this can be proved by reductio ad absurdum, and have a proof for that. However, I'm not sure how to prove this using any other method of proof. It's my first time taking a course in logic and I'm not sure where to go with this.
 A: Hint: Think about even and odd numbers. Can all 3 variables be odd?
A: Another solution of this question.
Let show by induction that the relation among three sizes of the right triangle is $$k,\frac{k^2-1}{2},\frac{k^2+1}{2}$$ respectively when $k$ is odd 
$$2k,k^2-1,k^2+1$$ respectively when $k$ is even
A: It can also be proven directly from a certain formula. The numbers $a$, $b$ and $c$ such that $a^2 + b^2 = c^2$ are called "Pythagorean triples." Given two integers $m$ and $n$ satisfying $0 < m < n$, you can obtain a Pythagorean triple with this "recipe":


*

*$a = n^2 - m^2$

*$b = 2mn$

*$c = m^2 + n^2$


So $a$ and $c$ could be odd, but $b$ is even, and that's enough for $abc$ to be even as well. For example: $5^2 + 12^2 = 13^2$ and $5 \times 12 \times 13 = 780$.
Maybe you distrust the formula above. That's fine. It might be possible for both $a$ and $b$ to be odd. But then $a^2 + b^2$ is even. For example: $3^2 + 5^2 = 34$ (which is even, though it's not itself a square).
A: We know , a,b,c are integers.
Now some basic principles:
odd number +another odd number=even number.
Odd + even = odd.
even+even = even .
These 3 relations hold true for all integers.
We also know that square of and odd integer is odd, similarly square of an even integer is even.
So now if see the equation  a²+b²=c², we can conclude:
*If both 'a' and 'b' are odd , then the above sum is even , therefore 'c' must be even.
*If one of 'a' and 'b' is odd and other is even, then the sum a²+b² is odd which means 'c' must be odd.
*If both 'a' and 'b' are even then 'c' must be even .
These 3 cases cover all the possibilities .
Now in any of these case, atleast 1 of the three integers is even and at most 2 can be odd, which means that the product 'abc' is ALWAYS EVEN.
Thank you.
--Class 12th student as of 2020
A: Suppose $a$, $b$, and $c$ are odd. Then $a=2p-1$, $b=2q-1$, and $c=2r-1$. Inserting these into our formula, $a^2+b^2=c^2$, we get $(2p-1)^2+(2q-1)^2=(2r-1)^2$. Expanding, $4p^2-2p+1+4q^2-2q+1=4r^2-2r+1$, or $4p^2-2p+4q^2-2q+1=4r^2-2r$.
When written as $2(2p^2-p+2q^2-q+1)-1$, we see that the LHS is odd. Likewise, we see that the RHS is even. Because we can never have a value that is both even and odd, we get a contradiction. We can then say one of $a$, $b$, or $c$ must be even.
Let's say that $a$ is even; then $a=2n$, and $abc=2nbc$. We see that $abc$ is even. The other two cases are similar.
Therefore, we can say that $abc$ is even, in general.
