169-follower problem

169 is a square and it is a sum of squares (\(169 = 12^2 + 5^2\)).

But its square can also be written as the sum of two squares. Let \[\large 169^2 = a^2 + b^2\] with \(a > b > 0\) and \(\gcd(a,b)=1\). Without using a calculator, find \(a + b\).

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