A positive integer

\[n\]

cannot be expressed as the sum of three squares if it is of the form \[4^n(8m+7)\]

.Proof

The squares modulo 8 are 0,1, 4 so a sum of three squares can be congruent modulo 8 to any of the numbers 0, 1, 2, 3, 4, 5, 6 but not 7. Hence no number of the form

\[8m+7\]

can be a sum of three squares.Suppose for some

\[n \gt 1, \; m \ge 0\]

we have \[4^n(8m+7)=x^2+y^2+z^2\]

.The left hand side of the last expression is congruent modulo 4 to 0 and as squares modulo 4 are either 0 or 1, it has to be the case that

\[x, \; y, \; z\]

are all even. Put \[x=2x_1, \; y=2y_1. \; z=2z_1\]

we get \[4^{n-1}(8m+7)=x_1^2+y_1^2+z_1^2\]

.If

\[n-1 \gt 1\]

then \[x_1, \; y_1, \; z_1\]

are all even and we can write \[4^{n-2}(8m+7)=x_2^2+y_2^2+z_2^2\]

.We can continue in this way, eventually expressing

\[8m+7\]

as a sum of three square - a contradiction so the assumption that \[4^n(8m+7)\]

can be written as a sum of three squares is false.