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The Problem


Rearranging the equation gives us:

a=r2−23.

We will analyze three cases for r: r=3n, r=3n+1, and r=3n+2, where n∈N.

(a) For r=3n:

r2−2=(3n)2−2=9n2−2.
This expression is not divisible by 3, indicating that a solution with r=3n is not feasible.

(b) For r=3n+1:

r2−2=(3n+1)2−2=9n2+6n+1−2=9n2+6n−1.
This expression is also not divisible by 3, so a solution with r=3n+1 is not possible.

(c) For r=3n+2:

r2−2=(3n+2)2−2=9n2+6n+4−2=9n2+6n+2.
This expression remains not divisible by 3, thus a solution with r=3n+2 is also not possible.

After examining all potential cases, we conclude that there are no integer solutions.

Note: While the equation r2=3a−2 does have integer solutions for r=3n+1 and r=3n+2, it does not have solutions for r=3n.