Posted November 28, 2013 whats the smallest number that can be expressed as the sum of 3 squares in 3 unique ways 0 Share this post Link to post Share on other sites

0 Posted November 29, 2013 A bit cheating, but one candidate would be 125= 0^{2} + 5^{2} + 10^{2 } = 3^{2} + 4^{2} + 10^{2} = 5^{2} + 6^{2} + 8^{2} 0 Share this post Link to post Share on other sites

0 Posted November 29, 2013 54 = 1^{2}+2^{2}+7^{2} = 2^{2}+5^{2}+5^{2} = 3^{2}+3^{2}+6^{2} 0 Share this post Link to post Share on other sites

0 Posted November 29, 2013 -125 = 0^{2} + (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.} ^{} -1 Share this post Link to post Share on other sites

0 Posted November 29, 2013 The answer depends whether you accept 0 as a square and whether you accept the same number twice. -1 Share this post Link to post Share on other sites

0 Posted November 29, 2013 0^{2 + }0^{2 }+ 0^{2} = (sqrt(-1)^{2} + (1^{2}) + 0^{2} = 2^{2}+2^{2}+(sqrt(-8))^{2}=0 1 Share this post Link to post Share on other sites

0 Posted November 29, 2013 101 = 1^{2}+6^{2}+8^{2} = 2^{2}+4^{2}+9^{2} = 4^{2}+6^{2}+7^{2} 0 Share this post Link to post Share on other sites

0 Posted December 2, 2013 (edited) @ 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 0 Share this post Link to post Share on other sites

0 Posted December 2, 2013 41 = 0^{2}+4^{2}+5^{2} = 1^{2}+2^{2}+6^{2} = 3^{2}+4^{2}+4^{2} 1 Share this post Link to post Share on other sites

0 Posted December 2, 2013 65 = 0^{2}+1^{2}+8^{2} = 0^{2}+4^{2}+7^{2} = 2^{2}+5^{2}+6^{2} 0 Share this post Link to post Share on other sites

0 Posted December 2, 2013 @PerhapsCheckITAgain I see no requirement that the numbers be square numbers just that i square three numbers -1 Share this post Link to post Share on other sites

0 Posted December 2, 2013 Barring complex numbers and limiting ourselves to integers... 0^{2}+1^{2}+1^{2}=0^{2}+1^{2}+(-1)^{2}=0^{2}+(-1)^{2}+(-1)^{2}=2 -1 Share this post Link to post Share on other sites

0 Posted December 2, 2013 i like this anwer the best though i was aiming for non zero answers. -1 Share this post Link to post Share on other sites

0 Posted December 2, 2013 (edited) @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 0 Share this post Link to post Share on other sites

0 Posted December 2, 2013 (edited) Barring complex numbers and limiting ourselves to integers... 0^{2}+1^{2}+1^{2}=0^{2}+1^{2}+(-1)^{2}=0^{2}+(-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 0 Share this post Link to post Share on other sites

0 Posted December 2, 2013 i like this anwer the best though i was aiming for non zero answers. phil1882, what Prime offered does not count as I explained. 0 Share this post Link to post Share on other sites

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

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