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.
I have parts of this, not enough. It is sometimes the case that problems are proposed for which the organizers do not genuinely possess a full proof.
Just so you know, it is quite difficult to find solutions with three distinct numbers. Meanwhile, the "new m" is the result of what is often called Vieta Jumping. Maybe I should put a little detail, as Carl may have finished the problem, but it depends on compatibility of the "jumping" versions. The original equation squares to $$ m^2 + n^2 + p^2 +2mn+2np+2pm - 4mnp -2m-2n-2p+1=0. $$ Collecting this as a quadratic equation in $m$ alone, we get $$ m^2 - (4np-2n-2p+2)m + (n+2 + 2np + p^2 -2n -2p+1 ) = 0 $$ If there is one integer solution with this $m,$ there is another we can call $m',$ with $m+m' = (4np-2n-2p+2),$ finally the new $m$ value is $$ m' = (4np-2n-2p+2) - m $$ The jump is from the triple (solution ) $(m,n,p)$ to $(m',n,p)$
I should include the reason the this version of jumping preserves squares, so also preserves non-squares. Now, if $m$ is a square, the fact that $\sqrt {mnp}$ is rational means $np$ is also a square. Note that we have $2 \sqrt{mnp} = m+n+p-1$ and $4 \sqrt{mnp} = 2m+2n+2p-2.$
Consider $$ \color{red}{Q = \left( \sqrt m - 2 \sqrt {np} \right)^2} \; . $$ $$ Q = m + 4np - 4 \sqrt{mnp} = m + 4np - (2m+2n+2p-2) \; , $$ $$ Q = 4np -2n-2p+2 - m \; . $$ This is the same as $m'$ in the jump. So, when $m$ is a square, $$ \color{red}{m' = \left( \sqrt m - 2 \sqrt {np} \right)^2} \; $$ is also a square.
m + n + p - 1 = 2 sqrt( m n p )
new m = 4 n p - 2 n - 2 p + 2 - m
(1,n,n) is a solution, maps to
( (2 n - 1)^2, n, n).
DISTINCT m > n > p :
50 9 2 new m: 2 9 2
243 25 3 new m: 3 25 3
289 50 2 new m: 9 50 2
676 49 4 new m: 4 49 4
1445 81 5 new m: 5 81 5
1682 50 9 new m: 2 50 9
1682 289 2 new m: 50 289 2
2401 243 3 new m: 25 243 3
2646 121 6 new m: 6 121 6
4375 169 7 new m: 7 169 7
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 $(1,n,n)$.
