## prove the sum of two even perfect squares is not a perfect square

1. The problem statement, all variables and given/known data
For all natural numbers, a and b, if a and b are both even, then (a^2+b^2) is not a perfect square. (prove this)

2. Relevant equations

3. The attempt at a solution
I tried proving by contradiction and got (2s)^2 +(2t)^2 =k^2.
which translates to 4s^2 +4t^2=k^2.
I don't know how to form the contradiction from here. Is it even possible?

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 $$(a+b)^2=a^2+2ab+b^2$$

Mentor
 Quote by vinnie 1. The problem statement, all variables and given/known data For all natural numbers, a and b, if a and b are both even, then (a^2+b^2) is not a perfect square. (prove this)
Are you sure you have the problem correct as stated? As stated this is easily proven false by counterexample.

## prove the sum of two even perfect squares is not a perfect square

 Quote by D H Are you sure you have the problem correct as stated? As stated this is easily proven false by counterexample.
so I set up the negation, then we assume a and b are even and that a^2 +b^2 is a perfect square. Then subbing 4 for a and 6 for b, we get a contradiction?

 Mentor Obviously 52 is not a perfect square. The conjecture does not say that the sum of squares of some specific pair of even numbers is not a square number. The conjecture says that the sum of squares of every pair of even numbers is not a square number.

 Quote by D H Obviously 52 is not a perfect square. The conjecture does not say that the sum of squares of some specific pair of even numbers is not a square number. The conjecture says that the sum of squares of every pair of even numbers is not a square number.
true, but using four and six as counterexamples.....

 Mentor 42+62=16+36=52. and 52 is not a perfect square. 4 and 6 do not form a counterexample.

 Quote by D H 42+62=16+36=52. and 52 is not a perfect square. 4 and 6 do not form a counterexample.
a counterexample in the negation of the conjecture.

 Mentor Correct, and 52 is not a perfect square. 4 and 6 are consistent with the conjecture.
 we assume the negation of the conjecture. which is for all natural numbers, a and b, if a and b are both even, then (a^2 +b^2) IS a perfect square. if we use 4 and 6 as counterexamples we do not get a perfect square, so we have a contradiction.....
 or are you saying we don't need the negation, just provide 4 and 6 as counterexamples and be finished....?
 There is an error.
 or maybe we're supposed to prove the conjecture false by counterexample... using 6 and 8 perhaps.

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 Quote by vinnie or maybe we're supposed to prove the conjecture false by counterexample... using 6 and 8 perhaps.
Well, sure. It is false, isn't it?

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