Is At Least One of a or b Even in the Pythagorean Theorem?

[Instinctlol]

Homework Statement

If a,b and c are integers and a²+b²=c², then at least one of a and b is even.


2. Contradiction statement
There exist an integer a,b,c such that a²+b²=c² and a or b is odd

The Attempt at a Solution

I am not sure if my contradiction statement is correct because of this part, then at least one of a and b is even. I think it means 1 has to be even and the other could be even or odd. Let me know if my contradiction is right or wrong and point me in the right direction to start.

 

[Mark44]

The statement "at least one of a and b is even" is true if either a or b is even, or if both a and b are even. The opposite of this statement is "neither a nor b is even."

 

[Instinctlol]

So either a or b is odd would be correct?

 

[vela]

Suppose a=1 and b=2. Then the statement "either a or b is odd" is true because a is odd. The statement "at least one of a and b is even" is true because b is even. So "either a or b is odd" can't be the opposite of "at least one of a and b is even."

 

[Mark44]

So either a or b is odd would be correct?

If you'll read more closely, you'll see that I said neither a nor b is odd.

 

[Instinctlol]

If you'll read more closely, you'll see that I said neither a nor b is odd.

I reread your statement and it says both a and b are not even so they both must be odd?

 

[Dick]

I reread your statement and it says both a and b are not even so they both must be odd?

Yes, to prove the contradiction you would assume they are both odd. Think about mod 4.