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phil1882

perfect powers

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-125 = 02 + (sqrt(-25))2 + (sqrt(-100)2 = (sqrt(-9))2 + (sqrt(-16))2 + (sqrt(-100))2 = (sqrt(-25))2 + (sqrt(-36)2 + (sqrt(-64))2

so simply put any answer that can define a number using three squares has an equally negative solution. So whatever the biggest number we can make using three squares in three different ways naturally leads us to the smallest square number in three different ways.

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@ BMAD, your solutions do not count, because sqrt(-25), sqrt(-100), sqrt(-9), etc. are not integers.

A square (number) is the square of an integer and is necessarily nonnegative.

@ Grimbal, it is not a question of whether someone accepts 0 as a square (number). It is a fact.

Square numbers are the squares of integers. The phrase "in 3 unique ways" is ambiguous.

Edited by Perhaps check it again

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@PerhapsCheckITAgain

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

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Barring complex numbers and limiting ourselves to integers...


02+12+12=02+12+(-1)2=02+(-1)2+(-1)2=2
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@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

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Barring complex numbers and limiting ourselves to integers...

02+12+12=02+12+(-1)2=02+(-1)2+(-1)2=2

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

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