Ever so equable II

Suppose there does exist a right prism with positive integer edge length such that its volume $V$ , surface area $S$ , and total edge length $E$ are all numerically equal.

Then let the area of the base of the prism be $B$ , let the perimeter of the base be $P$ , let the number of sides of the base be $n$ , and let the length of the lateral sides be $h$ . Then the volume is $V = Bh$ , the surface area is $S = 2B + Ph$ , and the total edge length is $E = 2P + nh$ .

Setting $S = E$ and solving for $h$ gives $h = \frac{2(B - P)}{n - P}$ , and then setting $V = E$ and substituting $h$ and solving for $B$ gives $\frac{1}{2}(P + n \pm \sqrt{(n - P)(n + 3P)})$ .

For $B$ to be a real number, $(n - P)(n + 3P) > 0$ or $n > P$ . However, if all the edge lengths are at least a positive integer, $P > n$ . This is a contradiction, so there cannot exist a right prism with positive integer lengths such that $V = S = E$ .

Say the base of the prism is an $n$ -gon with area $b$ and perimeter $p$ , and say the height of the prism is $h$ .

$p,n,h$ are all positive integers; $b$ is positive but may or may not be an integer.

We have: Volume: $bh$

Surface area: $2b+ph$

Total edge length: $2p+nh$

First we prove that $h>2$ by considering volume and surface area:

$bh=2b+ph$

$b(h-2)=ph$

Since $ph$ is positive, we must have $h>2$ .

Now, equating surface area and total edge length, we have

$2b+ph=2p+nh$

Multiplying both sides by $h$ :

$2bh+ph^2=2ph+nh^2$

The expressions for volume and total edge length give $bh=2p+nh$ ; substituting:

$2(2p+nh)+ph^2=2ph+nh^2$

Tidying and collecting terms:

$4p+2nh-2ph=nh^2-ph^2$

$4p=(h^2 - 2h)(n-p)$

The left-hand side is positive, so the right-hand side must be as well. But $h>2$ , so $h^2-2h$ is positive; hence $n-p$ is positive as well, ie $n>p$ .

So the base of the prism is an $n$ -gon with perimeter $p<n$ ; so at least some of its sides must have length less than $1$ . But then they can't all be positive integers; hence no such prism exists.