I2.2 If $b$ and $h$ are positive integers with $b<h$ and $b^{2}+h^{2}=b(a+h)+a h$, find the value of $b$.
$b<h$ and $b^{2}+h^{2}=b(2+h)+2 h$
$$
b^{2}+h^{2}=2 b+b h+2 h
$$
$(b+h)^{2}-2(b+h)=3 b h<3\left(\frac{b+h}{2}\right)^{2}$
(G.M. < A.M., given that $b<h$ )

Let $t=b+h, t^{2}-2 t<\frac{3 t^{2}}{4} \Rightarrow t^{2}-8 t<0$, where $t$ is a positive integer
$t-8<0 \Rightarrow t<8 \Rightarrow b+h<8 \Rightarrow 2 b<b+h<8 \Rightarrow 2 b<8 \Rightarrow b<4$
$b=1,2$ or 3
When $b=1,1+h^{2}=2+h+2 h \Rightarrow h^{2}-3 h-1=0 \Rightarrow h$ is not an integer, rejected
When $b=2,4+h^{2}=4+2 h+2 h \Rightarrow h^{2}-4 h=0 \Rightarrow h=4$
When $b=3,9+h^{2}=6+3 h+2 h \Rightarrow h^{2}-5 h+3=0 \Rightarrow h$ is not an integer, rejected $\therefore b=2$
Method $2 h^{2}-(b+2) h+b^{2}-2 b=0$
$\Delta=(b+2)^{2}-4\left(b^{2}-2 b\right)=m^{2}$, where $m$ is an integer
$-3 b^{2}+12 b+4=m^{2}$
$-3(b-2)^{2}+16=m^{2}$
$m^{2}+3(b-2)^{2}=16$, both $b$ and $m$ are integers
$m=0$, no integral solution for $b$
$m=1$, no integral solution for $b$
$m=2, b=4, h^{2}-6 h+8=0 \Rightarrow h=2$ or $h=4$, contradicting $b<h$, reject
$m=3$, no integral solution for $b$
$m=4, b=2, h=4$ (accept)