In a triangle $A B C$, with usual notations, if $a=5, b=4, \cos (A-B)=\frac{31}{32}$, then $c=$

Given: In a triangle, sides $a = 5$, $b = 4$, and $\cos(A - B) = \frac{31}{32}$. Find side $c$.

Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$, express the trigonometric functions in terms of the sides via the cosine rule:

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{16 + c^2 - 25}{8c} = \frac{c^2 - 9}{8c}$

$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{25 + c^2 - 16}{10c} = \frac{c^2 + 9}{10c}$

Using the sine rule, $\sin A = \frac{a \sin C}{c} = \frac{5 \sin C}{c}$ and $\sin B = \frac{b \sin C}{c} = \frac{4 \sin C}{c}$.

Substitute into the identity:

$\frac{31}{32} = \left(\frac{c^2 - 9}{8c}\right)\left(\frac{c^2 + 9}{10c}\right) + \left(\frac{5 \sin C}{c}\right)\left(\frac{4 \sin C}{c}\right) = \frac{c^4 - 81}{80c^2} + \frac{20 \sin^2 C}{c^2}$

Now express $\sin^2 C$ in terms of $c$ using $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{41 - c^2}{40}$:

$\sin^2 C = 1 - \left(\frac{41 - c^2}{40}\right)^2 = \frac{1600 - (41 - c^2)^2}{1600}$

Substitute back:

$\frac{31}{32} = \frac{c^4 - 81}{80c^2} + \frac{1}{c^2} \cdot \frac{20[1600 - (41 - c^2)^2]}{1600} = \frac{c^4 - 81}{80c^2} + \frac{1600 - (41 - c^2)^2}{80c^2}$

Combine terms:

$\frac{31}{32} = \frac{c^4 - 81 + 1600 - (1681 - 82c^2 + c^4)}{80c^2} = \frac{82c^2 - 162}{80c^2} = \frac{41c^2 - 81}{40c^2}$

Cross-multiply:

$31 \cdot 40c^2 = 32(41c^2 - 81)$
$1240c^2 = 1312c^2 - 2592$
$72c^2 = 2592$
$c^2 = 36$

Thus, $c = 6$.

Solution: $\boxed{6}$