Đáp án:
Giả sử $a;b;c \quad \not \vdots \quad 3$
$→$ $\left\{\begin{matrix} a^2≡1(mod\quad 3) & \\ b^2≡1(mod\quad 3) & \end{matrix}\right.$
$→a^2+b^2\quad ≡ \quad 2(mod\quad 3)$ nhưng $c^2≡1(mod\quad 3)$
mà $a^2+b^2=c^2→$ Mẫu thuẫn
$→abc \quad \vdots \quad 3$
Giả sử $a;b;c \quad \not \vdots \quad 4$
$→$ $\left\{\begin{matrix} a^2≡1(mod\quad 4) & \\ b^2≡1(mod\quad 4) & \end{matrix}\right.$
$→a^2+b^2\quad ≡ \quad 2(mod\quad 4)$nhưng $c^2≡1(mod\quad 4)$
mà $a^2+b^2=c^2→$ Mẫu thuẫn
$→abc \quad \vdots \quad 4$
Giả sử $a;b;c \quad \not \vdots \quad 5$
\(\left[ \begin{array}{l}a^2≡1(mod\quad 5)\\a^2≡-1(mod\quad 5); a^2≡4(mod\quad 5)\end{array} \right.\)
$→$ $abc \quad \vdots \quad 5$
Vì $(3;4;5)=1→abc\quad \vdots \quad 60$
