Getting from $m^2 + n^2 + p^2 = mnp + 4$ to $(m + n + p + 2)^2 = (m + 2)(n + 2)(p + 2)$

Question

$$m^2 + n^2 + p^2 = mnp + 4 \tag1$$

Adding $2(mn + np + pm)$ to both sides yields

$$(m + n + p)^2 = mnp + 2(mn + np + pm) + 4 \tag{2}$$

Adding now $4(m + n + p) + 4$ to both sides gives $$(m + n + p + 2)^2 = (m + 2)(n + 2)(p + 2) \tag{3}$$

I just want an explanation for $(3)$. Can it be verified by expanding? Is there another way of arriving at that last statement?

Also a link to similar questions is also welcome.

Answer 1

$\begin{align}((m+n+p)+2)^2 &= (m+n+p)^2 + 4(m+n+p) + 4 \\ &= mnp + 2(mn+np+pm)+4(m+n+p) + 8\\ & = (mnp + 2mn)+(2np+4n) + (4p+8) + (4m+2pm) \\ &=mn(p+2)+2n(p+2)+4(p+2)+2m(p+2)\\ &=(p+2)(2m+mn+2n+4)\\ & = (p+2)(2m(n+2)+2(n+2))\\ &=(m+2)(n+2)(p+2)\end{align}$

_{In the second line, $(m+n+p)^2= mnp+2(m+n+p)+4$ is used.}