アルキメデスの牛の問題
W = ( 1 2 + 1 3 ) B + Y B = ( 1 4 + 1 5 ) D + Y D = ( 1 6 + 1 7 ) W + Y w = ( 1 3 + 1 4 ) ( B + b ) b = ( 1 4 + 1 5 ) ( D + d ) d = ( 1 5 + 1 6 ) ( Y + y ) y = ( 1 6 + 1 7 ) ( W + w ) W + B = p 2 Y + D = q ( q + 1 ) 2 {\displaystyle {\begin{aligned}W&=\left({\frac {1}{2}}+{\frac {1}{3}}\right)B+Y\\B&=\left({\frac {1}{4}}+{\frac {1}{5}}\right)D+Y\\D&=\left({\frac {1}{6}}+{\frac {1}{7}}\right)W+Y\\w&=\left({\frac {1}{3}}+{\frac {1}{4}}\right)(B+b)\\b&=\left({\frac {1}{4}}+{\frac {1}{5}}\right)(D+d)\\d&=\left({\frac {1}{5}}+{\frac {1}{6}}\right)(Y+y)\\y&=\left({\frac {1}{6}}+{\frac {1}{7}}\right)(W+w)\\W+B&=p^{2}\\Y+D&={\frac {q(q+1)}{2}}\end{aligned}}}
最後の2つの条件は、W + B が平方数であり、Y + D が三角数であることを示す。