Find all integers a, b, c where c is a prime number such that
a^b + c and
a^b  c are square numbers.
Source: Pedro Henrique O. Pantoja, Brazil
We have
a^b+c=x^2
a^bc=y^2
So y^2+2c=x^2, and since the difference is even, x and y have the same parity.
Then, without loss of generalisation:
(y+2n)^2=x^2, e.g. (n=1): 1+2=3 and 3^2=1^2=8, (n=4): 2+4=6, and 6^22^2= 32, etc.
But (y+2n)^2y^2 = 4n(n+y), so c=2n(n+y), no prime solutions. This solves the problem set.
There are nonprime solutions for y^2, a^3, x^2 in arithmetic progression:
25^2+c=25^3, 25^3+c=175^2, c=15000,... but is there a smaller solution?
Edited on February 3, 2018, 1:15 am

Posted by broll
on 20180203 01:12:06 