Find all primes p where 15p+16 is a perfect square

[stan]

Started to work on this problem to find all primes that satisfy 15p+16.
Tried going this way that [MATH]15p+16=a^2\Rightarrow 15p=a^2-16=(a-4)(a+4)[/MATH]. But don't know how to proceed and put this correctly. By trying I found that if p=3, then it works but what comes next, I don't know.

 

[Denis]

Find all primes p where 15p+16 is a perfect square

If p = 3, then 15p + 16 = 45 + 16 = 61 : NOT a perfect square.
WHY are you saying "it works"?

I can find only 2 cases:
p = 7 : 121 (11)
p = 23 : 361 (19)

 

[Harry_the_cat]

Interesting question!
Using Excel it seems that there are only the two solutions that Denis states ie p=7 or p=23. But how can we be sure?

Stan's already said that 15p=(a+4)(a-4) in other words (a+4)(a-4) is a prime multiple of 15.

If a+4 = 15 then a-4 = 7 so (a+4)(a-4) = 15 * 7 (ie p = 7 which is prime)
If a -4 = 15 the a+4 =23 so (a+4)(a-4) = 15 *23 (ie p = 23 which is prime)

Now if you let a+4 = 15k (ie some multiple of 15) then a-4 = 15k-8 and so (a+4)(a-4) = 15k * (15k-8) = 15 * (k(15k-8)) which is a composite (not prime) multiple of 15.
A similar argument applies to a-4=15k.

So for (a+4)(a-4) to be a prime multiple of 15, this may only happen when either (a+4) =15 or (a-4) =15
It works in both cases so there are two and only two solutions (as given above)ie when p=7 or p=23.

 

[Harry_the_cat]

Another thought!

But what if (a+4) is a prime multiple of 3 and (a-4)=5 then it might be possible for (a+4)(a-4) to be a prime multiple of 15. Hmmmm!

Well that's not possible is it? If a-4 =5 then a+4 =13 which is NOT a prime multiple of 3.

Similarly for the other 3 possibilities
1. (a+4) is a prime multiple of 5 and (a-4) =3 … not possible
2. (a-4) is a prime multiple of 3 and (a+4) =5 … not possible (by definition p is positive)
3. (a-4) is a prime multiple of 5 and (a+4) =3 … not possible (by definition p is positive)

So only the two solutions previously stated.