(2^(p-1)-1)/p is not a square for p >7 and prime

@jamestanton raised the following (implicit) question on Twitter on September 8, 2017:

In the spirit of playing around and not thinking too hard, just computing to see what might be going on, I wrote a short piece of Mathematica® code that will run through the primes $p$ from the $2^{nd}$ through the $1000^{th}$ and check when $(2^{p-1}-1)/p$ is a square:

n = 2;
L = {};
While[n <= 1000,
If[IntegerQ[Sqrt[(2^(Prime[n] – 1) – 1)/Prime[n]]],
L = {L, Prime[n]}]; n++]
L = Flatten[L]

with the result {3, 7}.

So no more primes $p >7$ for which $(2^{p-1}-1)/p$ is a square.

To continue in this vein I just let Mathematica® keep on running through the primes greater than 7, stopping only if it finds a prime $p >7$ for which $(2^{p-1}-1)/p$ is a square:

n = 5;
While[IntegerQ[Sqrt[(2^(Prime[n] – 1) – 1)/Prime[n]]] == False, n++]
n
Prime[n]

I kept this running, with no positive result and aborted at which point I found $n = 43984$ , and the $n^{th}$ prime to be $532159$ .

So by now I was thinking that maybe $p = 3$ and $p = 7$ is it for primes $p$ for which $(2^{p-1}-1)/p$ is a square.

If that is indeed the case, then it would be sort of nice to figure where in the non-square numbers $(2^{p-1}-1)/p$ appears.

Now, the $n^{th}$ non-square number is $\lfloor \sqrt{n} + n + 1/2 \rfloor$ .

As a reminder: $\lfloor x \rfloor$ is the greatest integer $\leq x$ .

You can find this fact, for example, in Jim Tanton’s book Mathematics Galore! in Chapter 4.

So the idea is to find an integer $n$ for which $(2^{p-1}-1)/p =\lfloor \sqrt{n} + n + 1/2 \rfloor$ when $p > 7$ is prime.

If we drop the floor function temporarily we get a quadratic equation in $n$ and the smaller of the roots – which a bit of reflection tells us is the correct root – is $(2^{p-1}-1)/p -\sqrt{(2^{p-1}-1)/p -1/4}$ .

Of course this root is NOT an integer, and again a bit of reflection tells us that we want the ceiling of this root, so we set $n:= \lceil(2^{p-1}-1)/p -\sqrt{(2^{p-1}-1)/p -1/4} \rceil$ .

So now the idea is to prove that if $p >7$ is prime and

$n:= \lceil(2^{p-1}-1)/p -\sqrt{(2^{p-1}-1)/p -1/4} \rceil$

then

$(2^{p-1}-1)/p =\lfloor \sqrt{n} + n + 1/2 \rfloor$

This would tell us that $(2^{p-1}-1)/p$ is non-square for $p >7$ , and gives us something concrete with which to work.