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 A couple of squares (Posted on 2018-02-02)
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

 See The Solution Submitted by Ady TZIDON No Rating

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 Possible solution | Comment 5 of 9 |

We have
a^b+c=x^2
a^b-c=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 6^2-2^2= 32, etc.
But (y+2n)^2-y^2 = 4n(n+y), so c=2n(n+y), no prime solutions. This solves the problem set.

There are non-prime 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 2018-02-03 01:12:06

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