At first note that for any prime $p$ of the form $4k+1$ and any negative integer $a$ there is a residue $x\ (mod p^a)$ such that $x^2+1\equiv 0\ (mod\ p^a)$.Indeed, for $ a=1$ it is true.We can climb on higher powers of $p$ by Hensel's lemma. Now note that $c^2+1$ can't be a prime otherwise it will devide at least one of $a^2+1$ or $b^2+1$.So it must be composite. If $c^2+1$ is even note that $4$ can't devide $c^2+1$.So let $c=2p_1^{a_1}p_2^{a_2}...p_r^{a_r}$ where $p_i's$ are prime of the form $4k+1$.Take $u=1$.Now we claim that there exists even $x$ such that $x<p_1^{a_1}p_2^{a_2}...p_r^{a_r}=(c^2+1)/2<c^2$ such that $x^2+1\equiv 0\ (mod\ p_1^{a_1}p_2^{a_2}...p_r^{a_r})$.Indeed there is $x_i^2\equiv -1\ (mod\ p_i^{a_i})$ for all $i=1,2,...,r$by Hensel and so by CRT there is $x (mod\ p_1^{a_1}p_2^{a_2}...p_r^{a_r})$ which satisfy the equation.Now if $x$ is even we take $v=x$ otherwise take $v=p_1^{a_1}p_2^{a_2}...p_r^{a_r}-x$.Take $u=1$.Ofcourse $(u,v,c)$ is a interesting triple. Now let $c^2+1= odd=p_1^{a_1}p_2^{a_2}...p_r^{a_r}$ where all the primes are of the form $4k+1$.Now at least one prime divisior of $c^2+1$ will be less than $c$.Let this prime be $p_1$.There is $x<p_1<c$ such that $x^2+1\equiv 0\ (mod\ p_1)$. Now we claim that there exists $y<(c^2+1)/p_1<c^2$ such that $y^2+1\equiv 0( mod\ (c^2+1)/p_1)$ and $y^2+1\not=0\ (mod\ c^2+1)$. At first note that if $p^{a_1-1}|y^2+1$ and $p^{a_1}|y^2+1$ then $(y-p^{a_1-1})^2+1\not=0(mod\ p^{a_1})$ and $(y-p^{a_1-1})^2+1\equiv 0(mod\ p^{a_1-1})$ .And for all the other primes we can take $y_i^2+1\equiv 0( mod\ p^{a_i})$ for all $i=2,3,...r$.So by CRT the claim is proved. Now take $u=x$ and $v=y$.We get $(u,v,c)$ a interesting triple.