Algebra divisibility proof How can i show that if $24a^2 +1 = b^2$ then $5$ divides $ab$??
I tried writing 24 as 25-1  but i did not get anything interesting 
I have $25a^2=a^2+b^2-1$
But from there I don't know what to do 
Any help would be appreciated
Thanks in advance!!
 A: Ignoring the fact that the equation has no solution in integers, $a$ can not be divisible by $5$ because this would imply $5$ divides $5b^2-24a^2=1$, a contradiction. However $b$ is divisible by $5$. For $5b^2= 24a^2+1 = 25a^2-(a^2-1)$ implies $5$ divides $a^2-1$. Hence $a$ is either $1$ or $-1$ modulo $5$. But in either case you will find $24a^2+1$ is divisible by $25$. Thus $25$ divides $5b^2$ which implies $5$ divides $b^2$ which implies $5$ divides $b$, because $5$ is a prime.
A: Actually, $24a^2 + 1 = 5 b^2$ is not possible in integers. 
480    factored   2^5 * 3 *  5

        1.             1          20         -20   cycle length             2
        2.            -1          20          20   cycle length             2
        3.             4          20          -5   cycle length             2
        4.            -4          20           5   cycle length             2
        5.             3          18         -13   cycle length             4
        6.            -3          18          13   cycle length             4
        7.             7          16          -8   cycle length             4
        8.            -7          16           8   cycle length             4

      form class number is   8

A: Here's a little fact:  If $x$ is not divisible by $5$ then when you divide $x^2$ by $5$ the remainder is either $1$ or $4$.  You can see this by noting that $x=5q+r$ for $r=1,2,3,4$ and squaring each case.
Now suppose that neither $a$ nor $b$ is divisible by $5$ and take your equation
$$25a^2 = a^2+b^2 -1.$$
The left side is a multiple of $5$, therefore so is the right side.  Now try plugging in $1$ and $4$ for $a^2$ and $b^2$ to see if you can get a multiple of $5$ on the right.  You can't.  So you have a contradictions which shows that one of $a$ or $b$ has to be a multiple of $5$.
