4. $a, b, c$ are all natural numbers greater than 20, one of them has an odd number of positive divisors, the other two each have exactly three positive divisors, and $a+b=c$. The smallest value of $c$ that satisfies the above conditions is . $\qquad$
4. Note first that a natural number has an odd number of positive divisors if and only if it is a perfect square. In particular, if a natural number has exactly 3 positive divisors, this number must be the square of a prime.
Let \(a=p^{2}, b=q^{2}, c=r^{2}\). From \(a+b=c\), we get
\[
p^{2}+q^{2}=r^{2} \text {. }
\]

where \(p, q, r\) are all positive integers, and \(p<r, q<r, p, q, r\) with at least two being prime numbers.
Further discussion leads to the conclusion that the minimum value of \(c\) is 169.