We have two cases: Case 1: $p>2$ By Fermat's Little Theorem we have that $m^2+n^2+1\equiv m^2+n^2+p+1\equiv n^{2p}\equiv n^2(mod p)$,so $m^2+1\equiv 0(mod p)$. Suppose that $p\equiv 3(mod 4)$.Since $p|m^2+1$ we would obtain $p|1$,contradiction!Therefoe $p\equiv 1(mod 4)$. We have $0\equiv n^{2p}-n^2\equiv m^2+p+1\equiv m^2+2(mod 4)$,thus $m^2\equiv 2(mod 4)$,contradiction! Therefore there are no soluntions in this case. Case 2: $p=2$ We have $n^4=m^2+n^2+3$.It's obvious that $m<n^2$,so $m\le n^2-1$.So $n^4=m^2+n^2+3\le (n^2-1)^2+n^2+3=n^4+4-n^2$,which is equivalent to $n^2\le 4$ i.e. $n\le 2$.Therefore $n\in \{0,1,2\}$. Cheking the cases $n=0$,$n=1$ and $n=2$ we conclude that $(m,n,p)=(3,2,2)$ is the only solution to the given equation.

I think that this problem is better than P4 2012.

huricane wrote: $(m,n,p)=(3,2,2)$ is the only solution to the given equation. No,the solution is a permutations of $(m,n,p)=(\mp 3,\mp 2,2)$

Does $p$ even need to be a prime? I managed to get the same result without using that piece of information and I am very confused.

huricane wrote: We have two cases: Case 1: $p>2$ By Fermat's Little Theorem we have that $m^2+n^2+1\equiv m^2+n^2+p+1\equiv n^{2p}\equiv n^2(mod p)$,so $m^2+1\equiv 0(mod p)$. Suppose that $p\equiv 3(mod 4)$.Since $p|m^2+1$ we would obtain $p|1$,contradiction!Therefoe $p\equiv 1(mod 4)$. We have $0\equiv n^{2p}-n^2\equiv m^2+p+1\equiv m^2+2(mod 4)$,thus $m^2\equiv 2(mod 4)$,contradiction! Therefore there are no soluntions in this case. Case 2: $p=2$ We have $n^4=m^2+n^2+3$.It's obvious that $m<n^2$,so $m\le n^2-1$.So $n^4=m^2+n^2+3\le (n^2-1)^2+n^2+3=n^4+4-n^2$,which is equivalent to $n^2\le 4$ i.e. $n\le 2$.Therefore $n\in \{0,1,2\}$. Cheking the cases $n=0$,$n=1$ and $n=2$ we conclude that $(m,n,p)=(3,2,2)$ is the only solution to the given equation. Another way to finish case 2 is rearrange to $$n^4-m^2=(n^2+m)(n^2-m)=n^2+3\implies n^2+3\geq n^2+m \implies m\leq 3$$(assuming $m$ positive, can do case work based on its magnitude

We will start with a Lemma: Lemma: If $x,y$ are integers and $p=3 (mod.4)$ a prime number such that $p|x^{2}+y^{2}$, then $p|x$ and $p|y$. Proof: If $p$ divides one of the integers $x^{2}, y^{2}$, then it will also divide the other, and since $p$ prime, then $p|x$ and $p|y$. Suppose now for the sake of contradiction that $p$ does not divites either $x^{2}$ or $y^{2}$. Then $gcd.(x,p)=gcd.(y,p)=1$. First note that $p|x^{2}+y^{2} \Rightarrow x^{2}=-y^{2} (mod.p)$. Also let $p=4k+3$, for a natural number $k$. Applying $FLT$ we have that: $1=x^{p-1}=x^{(4k+3)-1}=x^{4k+2}=(x^{2})^{2k+1}=(-y^{2})^{2k+1}=-y^{4k+2}=-y^{(4k+3)-1}=-y^{p-1}=-1 (mod.p) \Rightarrow$ $2=0(mod.p) \Rightarrow p|2 \Rightarrow p=2$, a contradiction. To our problem now: First we see that if $(m,n,p)$ is a solution, then $(-m,n,p),(m,-n,p),(-m-n,p)$ are also solution. WLOG. suppose that $m,n \ge 0$. - If $p=3 (mod.4)$, then working $mod.p$ on the equation and from $FLT$ we have that $n^{2}=m^{2}+n^{2}+1 (mod.p) \Rightarrow$ $\Rightarrow m^{2}+1=0 (mod.p)$, and from the Lemma, we take that $p|1$, clearly a contradiction. - If $p=2 (mod.4)$, then $p=2$. Letting $n^{2}=l$, we have the equation $l^{2}-l-m^{2}-3=0$, which has discriminate: $D=1-4(-m^{2}-3)=4m^{2}+13$, and to have integral solutions, the discriminate must be a perfect square. We have the equation $4m^{2}+13=r^{2} \Rightarrow (r-2m)(r+2m)=13$, so we have the following two cases: Case 1 $r+2m=13$ $r-2m=1$ Adding and substracting them we take that $2r=14, 4m=12 \Rightarrow r=7, m=3 \Rightarrow$ $\Rightarrow l=\displaystyle{\frac{1+7}{2}}=4 \Rightarrow n=2 \Rightarrow (m,n,p)=(3,2,2)$. Case 2 $r+2m=-1$ $r-2m=-13$ This has the same solutions as the system in Case 1. -If $p=1(mod.4)$, then working $mod.4$ on the equation we have that $(n^{p})^{2}=m^{2}+n^{2}+2 (mod.4)$. Note that $LHS=0,1(mod.4), RHS=0,2,3(mod.4) \Rightarrow LHS=RHS=0(mod.4)$ so for $LHS$ it must be $n^{2}=0(mod.4)$, but for $RHS$ it must be $n^{2}=1(mod.4)$, a contradiction. - If $p=0(mod.4)$, then $4|p$, which is clearly a contradiction. Conclusively: $(m,n,p)=(3,2,2),(-3,2,2),(3,-2,2),(-3,-2,2)$.

Isn't this for JBMO short list?

$i)p=2$ \[n^4-n^2-3=m^2\]If $n^2 >4$ then $(n^2-1)^2 <m^2=n^4-n^2-3<(n^2)^2$ So $|n| \leq 2$ We get $\boxed{(n,m,p)=(\mp2,\mp3,2)}$ $ii)p>2$ \[n^2(n^{2p-2}-1)=m^2+p+1\]$p|m^2+1 \implies p \equiv 1(mod 4)$ $\implies 4|m^2+2$ Contradiction