jmc

Setting $a=b=0,$ we get $$2f(0)=f(0{)}^2+1.$$Then $f(0{)}^2-2f(0)+1=(f(0)-1{)}^2=0,$ so $f(0)=1.$

Setting $a=1$ and $b=-1,$ we get $$f(0)+f(-1)=f(1)f(-1)+1,$$so $f(-1)(f(1)-1)=0.$ This means either $f(-1)=0$ or $f(1)=1.$

First, we look at the case where $f(1)=1.$ Setting $b=1,$ we get $$f(a+1)+f(a)=f(a)+1,$$so $f(a+1)=1.$ This means $f(n)=1$ for all integers $n.$

Next, we look at the case where $f(-1)=0.$ Setting $a=b=-1,$ we get $$f(-2)+f(1)=f(-1{)}^2+1=1.$$Setting $a=1$ and $b=-2,$ we get $$f(-1)+f(-2)=f(1)f(-2)+1,$$which simplifies to $f(-2)=f(1)f(-2)+1.$ Substituting $f(-2)=1-f(1),$ we get $$1-f(1)=f(1)(1-f(1))+1,$$which simplifies to $f(1{)}^2-2f(1)=f(1)(f(1)-2)=0.$ Hence, either $f(1)=0$ or $f(1)=2.$

First, we look at the case where $f(1)=0.$ Setting $b=1,$ we get $$f(a+1)+f(a)=1,$$so $f(a+1)=1-f(a).$ This means $f(n)$ is 1 if $n$ is even, and 0 if $n$ is odd.

Next, we look at the case where $f(1)=2.$ Setting $b=1,$ we get $$f(a+1)+f(a)=2f(a)+1,$$so $f(a+1)=f(a)+1.$ Combined with $f(1)=2,$ this means $f(n)=n+1$ for all $n.$

Thus, there a total of $\boxed{3}$ functions: $f(n)=1$ for all $n,$ $f(n)=n+1$ for all $n,$ and $$f(n)=\{\begin{array}{cl}1& \text{if n is even},\\ 0& \text{if n is odd}.\end{array}$$We check that all three functions work.