I write $a=d_1,b=d_2$ and let them be relative prime and both odd, just like $k$ and $n$. We have to find $\infty$ many $a,b$ such that $a+b-k=n$, so $ab|(a+b-k)^2+1$. Obviously $a=b=1$ satisfy (but $n$ isn't positive yet). $a^2-[(k^2+3-4k)b+2k]a+[ b^2+k^2+1-2kb]=0$ Starting with $a=b=1$ as a solution, we can do the following: Now take solutions $(a,b) $ such that $b \ge a$, then $a'= (k^2+3-4k)b+2k-a >b$ and $(b,a')$ is another solution with $b < a'$. We can repeat this and from some time $a+b >k$ and hence $n=a+b-k$ will be positive, so we get $\infty$ many $n$.