120. If the numbers $a^{5}$ and $b^{5}$ ( $a>0$ and $b>0$ ) have the same units digit, then $a-b$ is divisible by 10. Find the numbers $a$ and $b$, if $a^{2}-b^{2}=1940$. Under the same conditions, find $a$ and $b$, if $a^{2}-b^{2}=1920$.
120. Since $a^{5}$ and $b^{5}$ end with the same digit as $a$ and $b$, $a-b$ ends with zero, hence $a-b$ is divisible by 10. From the system of equations $\left\{\begin{array}{l}a^{2}-b^{2}=1940 \\ a-b=10 k\end{array}\right.$ we find $a=\frac{97}{k}+5 k, b=\frac{97}{k}-5 k$. Since $b>0$, then $k<\sqrt{19.4}, k=1 ; 2 ; 3 ; 4 ;$ but $\frac{97}{k}$ must be an integer, so $k=1, a=102, b=92$.

Solving the system $\left\{\begin{array}{l}a^{2}-b^{2}=1920 \\ a-b=10 k\end{array}\right.$, we find $a_{1}=101$,
$b_{1}=91, \quad a_{2}=58, \quad b_{2}=38, \quad a_{3}=47, \quad b_{3}=17, \quad a_{4}=44$, $b_{4}=4$.