15. Let $a, b, c, d, m, n$ be integers, $a^{2}+b^{2}+c^{2}+d^{2}=1989, a+b+c+d$ is determined (and prove) the values of $m, n$.
15. By Cauchy-Schwarz inequality,
\[ a + b + c + d \leqslant 2 \sqrt{1989}a^{2} + b^{2} + c^{2} + d^{2} \]

it follows that \( m^{2} = 49 \) or 81.
Assume without loss of generality that \( a \leqslant b \leqslant c \leqslant d = n^{2} \).
If \( m^{2} = 49 \), then
\[ 
\begin{aligned}
(49 - d)^{2} & = (a + b + c)^{2} > a^{2} + b^{2} + c^{2} \\
& = 1989 - d^{2}.
\end{aligned}
\]

Thus,
\[ 
d^{2} - 49 d + 206 > 0.
\]

This implies \( d > 44 \) or \( d \leqslant 4 \). But
\[ 
\begin{array}{l}
45^{2} > 1989 > d^{2} \Rightarrow d < 45.
\end{array}
\]

Therefore, \( m^{2} \neq 49 \). Hence, \( m^{2} = 81, m = 9 \), and
\[ 
d = n^{2} \in \{25, 36\}.
\]

If \( d = n^{2} = 25 \), let \( a = 25 - p, b = 25 - q \), \( c = 25 - r, p, q, r \geqslant 0 \), then from the given conditions we derive
\[ 
\begin{array}{l}
p + q + r = 19, \\
p^{2} + q^{2} + r^{2} = 439,
\end{array}
\]

which contradicts \( (p + q + r)^{2} > p^{2} + q^{2} + r^{2} \). Therefore, \( n^{2} = 36, n = 6 \).

Note: It is not difficult to deduce that \( a = 12, b = 15, c = 18 \), \( d = 36 \).