Hi guys! Some of you may already know me as 
aaaaaaaa, and I'm asking a question:

Solve the equation:

j² - (7c)² = 99

j and c are in N (natural numbers).

I know that j² - (7c)² = (j+7c)(j-7c), but the 
question is: Is there a way to solve this without 
trial-and-error and find out that j=50, c=7?

j² - (7c)² = 99

(j + 7c)(j - 7c) = 99

For that factorization of 99, the two factors,
(j + 7c) and (j - 7c) differ by 14c,
a multiple of 14

The only factors of 99 are 1, 3, 9, 11, 33, 99   

So there are only three ways to write 99 as the product 
of a larger natural number times a smaller natural number,
namely these three ways

99×1, 33×3 and 11×9.

Of the three ways only the first are factors that differ
by a multiple of 14, namely 98, since 98 = 14×7, so 
14c = 14×7 so 

c = 7

and since j² - (7c)² = 99,

j² - (7×7)² = 99
j² - 49² = 99
j² - 2401 = 99
j² = 2500

j = 50

Edwin