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# perfect powers

Best Answer superprismatic, 02 December 2013 - 02:30 AM

Spoiler for or perhaps you were thinking different squares

Go to the full post

15 replies to this topic

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Posted 02 December 2013 - 02:36 AM

@PerhapsCheckITAgain

I see no requirement that the numbers be square numbers just that i square three numbers

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### #12 Prime

Prime

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Posted 02 December 2013 - 05:48 AM

Barring complex numbers and limiting ourselves to integers...

Spoiler for another possibility

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Past prime, actually.

### #13 phil1882

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Posted 02 December 2013 - 10:33 AM

i like this anwer the best though i was aiming for non zero answers.

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### #14 Perhaps check it again

Perhaps check it again

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Posted 02 December 2013 - 10:06 PM

@PerhapsCheckITAgain

I see no requirement that the numbers be square numbers just that i square three numbers

The problem stated that they are "squares."  "Squares" in the integer sense *means* unambiguously

that the squares of integers are the only ones permitted, that is, 0, 1, 4, 9, 16, ...

Edited by Perhaps check it again, 02 December 2013 - 10:12 PM.

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### #15 Perhaps check it again

Perhaps check it again

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Posted 02 December 2013 - 10:16 PM

Barring complex numbers and limiting ourselves to integers...

Spoiler for another possibility

User Prime, your solution doesn't count, because it amounts the *same* solution repeated, that is, 0^2 + 1^2 + 1^2 = 2, that is,

0 + 1 + 1 = 2.

Your 2nd and 3rd expressions don't give different sums of squares than your 1st expression.

Edited by Perhaps check it again, 02 December 2013 - 10:18 PM.

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### #16 Perhaps check it again

Perhaps check it again

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Posted 02 December 2013 - 10:19 PM

i like this anwer the best though i was aiming for non zero answers.

phil1882, what Prime offered does not count as I explained.

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