See a similar problem here. We have $ f(n+1)+f(n+3)=f(n+5)f(n+7)-1375$ and $ f(n+3)+f(n+5)=f(n+7)f(n+9)-1375$. Thus $ f(n+5)-f(n+1)=f(n+7)(f(n+9)-f(n+5))$. But $ f(n+9)-f(n+5)=f(n+11)(f(n+13)-f(n+9))$ and hence $ f(n+5)-f(n+1)=f(n+7)f(n+11)(f(n+13)-f(n+9)$. Continuing this, we have \[ f(n+5)-f(n+1)=f(n+7)f(n+11)\ldots f(n+4k-1)(f(n+4k+1)-f(n+4k-3)), \]for any positive integer $ k$. If $ f(n+5)-f(n+1)$ does not equal $ 0$, then none of $ f(n+7),f(n+11),\ldots,f(n+4k-1),(f(n+4k+1)-f(n+4k-3))$ may equal $ 0$. But they do not equal to $ 1$, and hence $ f(n+7),f(n+11),\ldots,f(n+4k-1)\ge2,f(n+4k+1)-f(n-4k-3)\ge1$. So $ f(n+5)-f(n+1)\ge2^{k-1}$ for any positive integer $ k$, which is impossible. Hence $ f(n+5)-f(n+1)$ must equal to $ 0$, i.e., $ f(n+5)=f(n+1)$. So we have $ f(4)=f(8)=\ldots$, $ f(5)=f(9)=\ldots$, $ f(2)=f(6)=f(10)=\ldots$, $ f(3)=f(7)=f(11)=\ldots$. Let $ f(4)=a,f(5)=b,f(2)=c,f(3)=d$. We have $ a+c=ac-1375$ and $ b+d=bd-1375$. It is easy to solve these equations. $ f(0)$ and $ f(1)$ can be any non negative integer unequal to $ 1$.