
#1
Oct810, 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? 



#2
Oct810, 06:36 PM

P: 372

[tex](a+b)^2=a^2+2ab+b^2[/tex]




#3
Oct810, 07:27 PM

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#4
Oct810, 07:42 PM

P: 23

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



#5
Oct810, 08:02 PM

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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.




#6
Oct810, 08:05 PM

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#7
Oct810, 08:08 PM

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P: 14,440

4^{2}+6^{2}=16+36=52. and 52 is not a perfect square. 4 and 6 do not form a counterexample.




#8
Oct810, 08:12 PM

P: 23





#9
Oct810, 08:18 PM

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P: 14,440

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




#10
Oct810, 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..... 



#11
Oct810, 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....?




#12
Oct810, 08:31 PM

P: 23

There is an error.




#13
Oct810, 08:35 PM

P: 23

or maybe we're supposed to prove the conjecture false by counterexample...
using 6 and 8 perhaps. 



#14
Oct810, 09:53 PM

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