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

by vinnie
Tags: perfect, prove, square, squares
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vinnie
#1
Oct8-10, 06:18 PM
P: 23
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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ZioX
#2
Oct8-10, 06:36 PM
P: 371
[tex](a+b)^2=a^2+2ab+b^2[/tex]
D H
#3
Oct8-10, 07:27 PM
Mentor
P: 15,166
Quote Quote by vinnie View Post
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.

vinnie
#4
Oct8-10, 07:42 PM
P: 23
Prove the sum of two even perfect squares is not a perfect square

Quote Quote by D H View Post
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
#5
Oct8-10, 08:02 PM
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P: 15,166
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
#6
Oct8-10, 08:05 PM
P: 23
Quote Quote by D H View Post
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
#7
Oct8-10, 08:08 PM
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P: 15,166
42+62=16+36=52. and 52 is not a perfect square. 4 and 6 do not form a counterexample.
vinnie
#8
Oct8-10, 08:12 PM
P: 23
Quote Quote by D H View Post
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
#9
Oct8-10, 08:18 PM
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P: 15,166
Correct, and 52 is not a perfect square. 4 and 6 are consistent with the conjecture.
vinnie
#10
Oct8-10, 08:18 PM
P: 23
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
#11
Oct8-10, 08:20 PM
P: 23
or are you saying we don't need the negation, just provide 4 and 6 as counterexamples and be finished....?
vinnie
#12
Oct8-10, 08:31 PM
P: 23
There is an error.
vinnie
#13
Oct8-10, 08:35 PM
P: 23
or maybe we're supposed to prove the conjecture false by counterexample...

using 6 and 8 perhaps.
Dick
#14
Oct8-10, 09:53 PM
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Thanks
P: 25,228
Quote Quote by vinnie View Post
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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