Prove the sum of two even perfect squares is not a perfect square

[vinnie]

Homework Statement

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)

Homework Equations

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?

 

[ZioX]

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

 

[D H]

Homework Statement

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.

 

[vinnie]

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?

 

[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.

 

[vinnie]

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...

 

[D H]

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

 

[vinnie]

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.

 

[D H]

Correct, and 52 is not a perfect square. 4 and 6 are consistent with the conjecture.

 

[vinnie]

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...

 

[vinnie]

or are you saying we don't need the negation, just provide 4 and 6 as counterexamples and be finished...?

 

[vinnie]

There is an error.

 

[vinnie]

or maybe we're supposed to prove the conjecture false by counterexample...

using 6 and 8 perhaps.

 

[Dick]

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?