perfect powers

[phil1882]

whats the smallest number that can be expressed as the sum of 3 squares in 3 unique ways

[superprismatic]

65 = 0²+1²+8² = 0²+4²+7² = 2²+5²+6²

[Grimbal]

A bit cheating, but one candidate would be 125= 0² + 5² + 10² = 3² + 4² + 10² = 5² + 6² + 8²

[superprismatic]

54 = 1²+2²+7² = 2²+5²+5² = 3²+3²+6²

[BMAD]

-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.}

[Grimbal]

The answer depends whether you accept 0 as a square and whether you accept the same number twice.

[BMAD]

0^{2 +} 0² + 0² = (sqrt(-1)² + (1²) + 0² = 2²+2²+(sqrt(-8))²=0

[superprismatic]

101 = 1²+6²+8² = 2²+4²+9² = 4²+6²+7²

[Perhaps check it again]

@ 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 December 2, 2013 by Perhaps check it again

[superprismatic]

41 = 0²+4²+5² = 1²+2²+6² = 3²+4²+4²

[BMAD]

@PerhapsCheckITAgain

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

[Prime]

Barring complex numbers and limiting ourselves to integers...

0²+1²+1²=0²+1²+(-1)²=0²+(-1)²+(-1)²=2

[phil1882]

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

[Perhaps check it again]

@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 December 2, 2013 by Perhaps check it again

[Perhaps check it again]

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 December 2, 2013 by Perhaps check it again

[Perhaps check it again]

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

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