A real age problem!

Let the ages of Red and Sepia be $r$ and $s$ respectively, Then $s=r+17$ and that

$\begin{cases} 100r + s = 101r + 17 =\color{#3D99F6} m^2 & ...(1) \\ 100(r+13) + s+13 = 101r + 1330 = \color{#3D99F6} n^2 & ...(2) \end {cases} \small \color{#3D99F6} \text{ where }n > m \text{ are positive integers.}$

$\begin{aligned} (2)-(1): \quad n^2-m^2 & = 1313 \\ (n-m)(n+m) & = \color{#3D99F6} 13\times 101 & \small \color{#3D99F6} \text{Note that 13 and 101 are primes.} \end{aligned}$

Assuming $\begin{cases} n - m = 13 \\ n+m = 101 \end{cases} \implies m = 44 \implies m^2 = {\color{#3D99F6}19}36 \implies r = \boxed{\color{#3D99F6}19}$ .