Question #262188

Find a counterexample to the statement “every positive integer greater than 7 can be written as the sum of the squares of three (not necessarily unique) integers.”


1
Expert's answer
2021-11-09T17:09:08-0500

Legendre's three-square theorem states that a natural number can be represented as the sum of three squares of integers n=x2+y2+z2{\displaystyle n=x^{2}+y^{2}+z^{2}} if and only if nn is not of the form n=4a(8b+7){\displaystyle n=4^{a}(8b+7)}  for nonnegative integers aa and b.b.

The first numbers that cannot be expressed as the sum of three squares (i.e. numbers that can be expressed as n=4a(8b+7)){\displaystyle n=4^{a}(8b+7)}) are


n=40(8(0)+7)=7{\displaystyle n=4^{0}(8(0)+7)}=7

n=40(8(1)+7)=15>7{\displaystyle n=4^{0}(8(1)+7)}=15>7

n=40(8(2)+7)=23{\displaystyle n=4^{0}(8(2)+7)}=23


n=41(8(0)+7)=28{\displaystyle n=4^{1}(8(0)+7)}=28

n=40(8(2)+7)=23{\displaystyle n=4^{0}(8(2)+7)}=23

n=40(8(3)+7)=31{\displaystyle n=4^{0}(8(3)+7)}=31

n=40(8(4)+7)=39{\displaystyle n=4^{0}(8(4)+7)}=39

n=40(8(5)+7)=47{\displaystyle n=4^{0}(8(5)+7)}=47

n=40(8(6)+7)=55{\displaystyle n=4^{0}(8(6)+7)}=55

n=41(8(0)+7)=60{\displaystyle n=4^{1}(8(0)+7)}=60

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