Let p be a prime number such that 17p + 1 is a perfect square. How many prime numbers satisfy this condition?

Question

mathmath333 · Answer

i hope there is only one that is 
$\large 19$

ganeshie8 · Answer

start with $$17p+1 = n^2$$

ganeshie8 · Answer

send that $1$ to other side and factor..

mathmath333 · Answer

$17p=n^2-1\
17p=(n-1)(n+1)\
$

ganeshie8 · Answer

Yes since $17$ and $p$ are primes, we they must equal the factors on right hand side in some order

ganeshie8 · Answer

$$17=n-1, ~~p=n+1$$

or

$$17=n+1, ~~p=n-1$$

mathmath333 · Answer

$17=n-1,p=n+1$
$n=18,p=19$


$17=n+1,p=n-1$
$n=16,p=15$

ganeshie8 · Answer

yes  $15$ is not a prime so the only solution is $p=19$

mathmath333 · Answer

if i use the fact the every prime greater than 3 can be of form $6n\pm1$ 
and if i substitute it in the equation and then apply $(mod~~ 6)$ can i prove that

ganeshie8 · Answer

i feel that gives a more complicated expression..

mathmath333 · Answer

ok i think that will not work because the the converse of $6n\pm1$ is not true.

ganeshie8 · Answer

good point