$m+n+p-1=2\sqrt{mnp}$ in positive integers If $m,n,p \in N$ and $m+n+p-1=2\sqrt{mnp}$, prove that at least one of 
$m,n,p$ is a perfect square.
There is a duplicate here: Perfect square : $ m+n+p-2\sqrt{mnp}=1$, but I am very confused.
Now I have proven $\gcd(m,n,p)=1$
And I let $\gcd(m,n)=d_1$, $\gcd(m/d_1,p)=d_2$, and $\gcd(n/d_1,p/d_2)=d_3$
I have $d_1d_2m_2^{2}+d_1d_3n_2^{2}+d_3d_2p_2^{2}-1=2d_1d_2d_3m_2n_2p_2$, where $d_1d_2m_2^{2}=m$, $d_1d_3n_2^{2}=n$ and $d_3d_2p_2^{2}=p$
Then I don't know what to do.

Here's a partial characterization:
First, let $a=2m-1,b=2n-1,c=2p-1$. The equation reduces to
$$\frac{a+b+c+1}{2}=2\sqrt{\left(\frac{a+1}{2}\right)\left(\frac{b+1}{2}\right)\left(\frac{c+1}{2}\right)},$$
which after some manipulations becomes
$$2abc+1=a^2+b^2+c^2.$$
Writing in terms of $a$ we get
$$a^2-2abc+(b^2+c^2-1)=0.$$
We see that we can, via Vieta Jumping, go from the solution $(a,b,c)$ to $(2bc-a,b,c)$. As such, if $a>bc$ (or, for that matter, $b>ca$ or $c>ab$), we can find a solution with smaller sum. 

Claim. If $a,b,c$ are positive real numbers $\geq 1$ where $a<bc,b<ca,c<ab$, then
$$2abc+1>a^2+b^2+c^2.$$
Proof. Assume the contrary. As $a^2-2abc+(b^2+c^2-1)\geq 0$, and $a$ cannot be greater than the larger root of this quadratic (as that would give $a>bc$), we must have
$$a\leq bc-\sqrt{(b^2-1)(c^2-1)}.$$
However, $a>b/c,c/b$. WLOG let $b\geq c$. Then
$$\frac{b}{c}< bc-\sqrt{(b^2-1)(c^2-1)}$$
$$\sqrt{(b^2-1)(c^2-1)} < bc-\frac{b}{c}$$
$$b^2c^2-b^2-c^2+1 < b^2c^2-2b^2+\frac{b^2}{c^2}$$
$$(b^2-c^2)\left(1-\frac{1}{c^2}\right) < 0.$$
However, $b\geq c$ and $c\geq 1$ imply together that both factors on the left are positive, a contradiction.

From this claim, we see that we can reduce any solution for which $a\neq bc,b\neq ca,c\neq ab$ to one with a smaller sum. As a result, every solution can be "jumped to" from a solution where, WLOG, $a=bc$. This reduces to $(b^2-1)(c^2-1)$, so either $b=1$ or $c=1$, which results in our solution set $(1,n,n)$ (and permutations, of course). 
So, every solution can be "jumped to" from a solution of the form $(1,n,n)$.
