35 Find all positive integers $a$, $b$, and $c$, such that $a^{2}+1$ and $b^{2}+1$ are both prime, and satisfy
$$\left(a^{2}+1\right)\left(b^{2}+1\right)=c^{2}+1$$
35. Let's assume $a \leqslant b$, from the condition we have
$$a^{2}\left(b^{2}+1\right)=c^{2}+1-b^{2}-1=(c-b)(c+b)$$

Thus, $b^{2}+1 \mid c-b$ or $b^{2}+1 \mid c+b$ (here we use the fact that $b^{2}+1$ is a prime).

If
$$b^{2}+1 \mid c-b$$

then $\square$
$$c-b \geqslant b^{2}+1 \text { (note that } c>b \text { is obvious), }$$

i.e., $\square$
$$c \geqslant b^{2}+b+1$$

In this case,
$$c^{2}+1 \geqslant\left(b^{2}+b+1\right)^{2}+1>\left(b^{2}+1\right)^{2} \geqslant\left(a^{2}+1\right)\left(b^{2}+1\right)$$

Contradiction.
If
$$b^{2}+1 \mid c+b$$

then $\square$
$$c+b \geqslant b^{2}+1$$

i.e., $\square$
$$c \geqslant b^{2}-b+1$$

Thus,
$$\begin{aligned}
c^{2}+1 & \geqslant\left(b^{2}-b+1\right)^{2}+1 \\
& =\left(b^{2}+1\right)^{2}-2 b\left(b^{2}+1\right)+b^{2}+1 \\
& =\left(b^{2}+1\right)\left((b-1)^{2}+1\right)
\end{aligned}$$

Notice that, if $a=b$, then $c^{2}+1=\left(a^{2}+1\right)^{2}$, which cannot hold for positive integers $a$ and $c$ (since the difference of squares of two positive integers is at least 3), so $a<b$, i.e., $a \leqslant b-1$, thus
$$c^{2}+1 \geqslant\left(b^{2}+1\right)\left((b-1)^{2}+1\right) \geqslant\left(b^{2}+1\right)\left(a^{2}+1\right)$$

Combining with the condition, we have
$$a=b-1, c=b^{2}-b+1$$

At this point, since $a^{2}+1$ and $b^{2}+1$ are both primes, $b^{2}+1$ is odd, so $b$ is even, and thus $a=b-1$ is odd, and $a^{2}+1$ is even, so $a=1$, hence $b=2, c=3$. When $(a, b, c)=$ $(1,2,3)$ or $(2,1,3)$, the condition is satisfied, and these are the required solutions.