@jamestanton raised
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 from the
through the
and check when
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 for which
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 for which
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 , and the
prime to be
.
So by now I was thinking that maybe
and
is it for primes
for which
is a square.
If that is indeed the case, then it would be sort of nice to figure where in the non-square numbers appears.
Now, the non-square number is
.
As a reminder:
is the greatest integer
.
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 for which
when
is prime.
If we drop the floor function temporarily we get a quadratic equation in and the smaller of the roots – which a bit of reflection tells us is the correct root – is
.
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 .
So now the idea is to prove that if is prime and
then
This would tell us that is non-square for
, and gives us something concrete with which to work.