Primes and Squares: prove that 3 is only prime that's 1 less than square number

[apple2357]

A problem appeared in a newspaper today that made me think...


It was prove that 3 is the only prime which is one less than a square number.

The approach is reasonably straightforward:
One less than a square number can be expressed as n² -1 = (n-1)(n+1). For this to be prime n-1 or n+1 would have to equal 1 so n=2, so (n+1)=3.

I started thinking about 4 less than a square number and the same reasoning gives 5 which is 4 less than 9
Then 9 less than a square number would be 7 and 16
Or 25 less than a square number would be 11, 36

However, if i wanted 16 less than a square number... n²-16 = (n-4)(n+4) and the same reasoning would say n-4=1 , so n=5 which would give (n+4)=9, but this is not prime?? Why does the same reasoning fail here?
I genuinely don't know the answer yet! Any thoughts!

And is there more work done on this area? ( I notice that prime 7 is one less than a cube number!)

 

[Dr.Peterson]

A problem appeared in a newspaper today that made me think...


It was prove that 3 is the only prime which is one less than a square number.

The approach is reasonably straightforward:
One less than a square number can be expressed as n² -1 = (n-1)(n+1). For this to be prime n-1 or n+1 would have to equal 1 so n=2, so (n+1)=3.

I started thinking about 4 less than a square number and the same reasoning gives 5 which is 4 less than 9
Then 9 less than a square number would be 7 and 16
Or 25 less than a square number would be 11, 36

However, if i wanted 16 less than a square number... n²-16 = (n-4)(n+4) and the same reasoning would say n-4=1 , so n=5 which would give (n+4)=9, but this is not prime?? Why does the same reasoning fail here?
I genuinely don't know the answer yet! Any thoughts!

And is there more work done on this area? ( I notice that prime 7 is one less than a cube number!)

What you are proving goes only in one direction: not that IF n-k = 1 THEN p = n+k will be prime and p^2 = n^2-k^2, but that IF p is a prime such that p^2 = n^2-k^2, THEN p = n+k where n-k = 1 (that is, p = 2k + 1).

So there is no guarantee that the number you get in any of these cases is a prime; just that it can't be a prime unless you obtain it the way you did.

As for cubes, try using the formula n^3 - k^3 = (n - k)(n^2 + nk+ k^2).

 

[apple2357]

Ah yes. Makes good sense!