ヤコビの二平方定理
F ( 1 4 ) = π ( 2 q 1 / 4 ) 2 ( 2 q 1 / 4 ) 2 cos 2  π 4 ∏ m = 1 ∞ ( 1 − q 2 m ) 2 ( 1 + q 2 m ) 4 ( 1 − q 2 m ) ( 1 + i q 2 m − 1 ) ( 1 − i q 2 m − 1 ) ( 1 − q 2 m ) ( 1 − i q 2 m − 1 ) ( 1 + i q 2 m − 1 ) ( 1 − q 2 m ) 2 ( 1 + i q 2 m ) 2 ( 1 − i q 2 m ) 2 = 2 π ∏ m = 1 ∞ ( 1 − q 2 m ) 2 ( 1 + q 2 m ) 4 ( 1 + i q 2 m − 1 ) 2 ( 1 − i q 2 m − 1 ) 2 ( 1 + i q 2 m ) 2 ( 1 − i q 2 m ) 2 = 2 π ∏ m = 1 ∞ ( 1 − q 4 m ) 2 ( 1 + q 2 m ) 2 ( 1 + q 4 m − 2 ) 2 ( 1 + q 4 m ) 2 = 2 π ∏ m = 1 ∞ ( 1 − q 4 m ) 2 ( 1 + q 4 m ) 2 ( 1 + q 4 m − 2 ) 2 ( 1 + q 4 m − 2 ) 2 ( 1 + q 4 m ) 2 = 2 π ∏ m = 1 ∞ ( 1 − q 4 m ) 2 ( 1 + q 4 m − 2 ) 4 {\displaystyle {\begin{aligned}F\left({\tfrac {1}{4}}\right)&={\frac {\pi \left(2q^{1/4}\right)^{2}}{\left(2q^{1/4}\right)^{2}\cos ^{2}{\tfrac {\pi }{4}}}}\prod _{m=1}^{\infty }{\frac {(1-q^{2m})^{2}(1+q^{2m})^{4}(1-q^{2m})(1+iq^{2m-1})(1-iq^{2m-1})(1-q^{2m})(1-iq^{2m-1})(1+iq^{2m-1})}{(1-q^{2m})^{2}(1+iq^{2m})^{2}(1-iq^{2m})^{2}}}\\&=2\pi \prod _{m=1}^{\infty }{\frac {(1-q^{2m})^{2}(1+q^{2m})^{4}(1+iq^{2m-1})^{2}(1-iq^{2m-1})^{2}}{(1+iq^{2m})^{2}(1-iq^{2m})^{2}}}\\&=2\pi \prod _{m=1}^{\infty }{\frac {(1-q^{4m})^{2}(1+q^{2m})^{2}(1+q^{4m-2})^{2}}{(1+q^{4m})^{2}}}\\&=2\pi \prod _{m=1}^{\infty }{\frac {(1-q^{4m})^{2}(1+q^{4m})^{2}(1+q^{4m-2})^{2}(1+q^{4m-2})^{2}}{(1+q^{4m})^{2}}}\\&=2\pi \prod _{m=1}^{\infty }(1-q^{4m})^{2}(1+q^{4m-2})^{4}\\\end{aligned}}}
例えば、
r 2 ( 25 ) = 4 ( ( − 1 ) 1 − 1 2 + ( − 1 ) 5 − 1 2 + ( − 1 ) 25 − 1 2 ) = 12 {\displaystyle r_{2}(25)=4\left((-1)^{\frac {1-1}{2}}+(-1)^{\frac {5-1}{2}}+(-1)^{\frac {25-1}{2}}\right)=12}
であるが、実際に25を高々二個の平方数の和で表す方法は
25 = ( ± 5 ) 2 + 0 2 = 0 2 + ( ± 5 ) 2 = ( ± 4 ) 2 + ( ± 3 ) 2 = ( ± 3 ) 2 + ( ± 4 ) 2 {\displaystyle {\begin{aligned}25&=(\pm 5)^{2}+0^{2}\\&=0^{2}+(\pm 5)^{2}\\&=(\pm 4)^{2}+(\pm 3)^{2}\\&=(\pm 3)^{2}+(\pm 4)^{2}\\\end{aligned}}}
であり、符号と順序を区別すれば12個になる。