The Problem:
Suppose for a positive integer n both 5n+1 and 7n+1 are perfect squares. Show that n is divisible by 24

My attempt:
Since 5n+1 is a perfect square and n is a positive integer then, the n's for which 5n+1 is a perfect square are: n= 3, 7, 16, 24 ..........
and for 7n+1 : n = 5, 9, 24......
therefore the least common 'n' which makes both 5n+1 and 7n+1 a perfect square is 24 and therefore 'n' is divisible by 24.

Any better approach to this problem?

This only says that n = 24 works. It does not prove the general statement.

Thanks Alex! Got the point. Will try for a general solution.

A hint:

Try reducing the expressions mod 3 and mod 8. Then, what are all the squares in mod 3 and mod 8? From there, you can work out the details.

Many thanks! mod 3 and mod 8 works out beautifully!

You're welcome. I'm glad to hear that. It is sometimes difficult to give just the right amount of hint. Too little and there is frustration, too much and the fun is spoiled.