73. Find three numbers such that the sums of all three and each pair are squares.
73. Diophantus' Solution. Let the sum of all three numbers be $x^{2}+2 x+1=(x+1)^{2}$

Let $\mathrm{I}+\mathrm{II}=x^{2}$, then $\mathrm{III}=2 x+1$.

Let $\mathrm{II}+\mathrm{III}=(x-1)^{2}=x^{2}-2 x+1$, then I will be found by subtracting the sum of II + III, which is $x^{2}-2 x+1$, from the total sum $x^{2}+2 x+1$. We get $\mathrm{I}=4 x$. Therefore, $\mathrm{I}=x^{2}-4 x$. Hence, $\mathrm{I}+\mathrm{III}=6 x+1$ must be a square. Let $6 x+1=121$, then $x=20, \mathrm{I}=80 ; \mathrm{II}=320 ; \mathrm{III}=41$.