15. Let $f(x)=x^{2}+p x+q$. If the inequality $|f(x)|>2$ has no solution in the interval $[1,5]$. Try to find all real pairs $(p, q)$.

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15. First, let's assume such a pair of real numbers $(p, q)$ exists,

then $f(x)=x^{2}+p x+g$ for $x \in[1,5]$, has $|f(x)| \leqslant 2$. Take $x=1, 3, 5$.
$\Rightarrow|1+p+g| \leqslant 2$
(1), $|9+3 p+g| \leqslant 2$
(2), $|25+5 p+g| \leqslant 2$
(1) + (2) $\Rightarrow 4 \geqslant|1+p+g|+|9+3 p+g| \geqslant|8+2 p| \Rightarrow-6 \leqslant p \leqslant-2$.
(2) + (3) $\Rightarrow 4 \geqslant|9+3 p+g|+|25+5 p+g| \geqslant|16+2 p| \Rightarrow-10 \leqslant p \leqslant$
-6 . So $p=-6$.
Substitute into (1), (2) $\Rightarrow|g-5| \leqslant 2$. So $3 \leqslant g \leqslant 7$.
$|g-9| \leqslant 2$. So $7 \leqslant g \leqslant 11$. So $g=7$.
When $(p, q)=(-6,7)$, $f(x)=x^{2}-6 x+7=(x-3)^{2}-2$.
For $x \in[1,5]$, we have $-2 \leqslant f(x) \leqslant 2^{2}-2=2$. So $|f(x)| \leqslant 2$ holds.
In conclusion, the required pair of real numbers $(p, g)=(-6,7)$.