`A= (a^(2020) + b^(2019) + c^(2018) ) - (a^(2016) + b^(2015) + c^(2014))`
`=a^(2016)(a^4 - 1) + b^(2015)(b^4 - 1) +c^(2014)(c^4 - 1)`
`= a^(2016)(a^2+1)(a - 1)(a + 1) + b^(2015)(b^2+1)(b - 1)(b + 1) + c^(2014)(c^2+1)(c - 1)(c + 1)` `a(a - 1)(a + 1)` là tích` 3` số tự nhiên liên tiếp nên
` a(a - 1)(a + 1) \vdots 2 ; 3`
`⇒ a(a - 1)(a + 1) \vdots6`
`⇒ a^(2016)(a^2+1)(a - 1)(a + 1)`
Tương tự `b^(2015)(b^2+1)(b - 1)(b + 1) \vdots 6`
`c^(2014)(c^2+1)(c -1)(c + 1) \vdots 6`
`⇒= a^(2016)(a^2+1)(a - 1)(a + 1) + b^(2015)(b^2+1)(b - 1)(b + 1) + c^(2014)(c^2+1)(c - 1)(c +1)\vdots 6`
`⇒A\vdots 6 (1)`
mặt khác :
`A= a^(2016)(a^2+1)(a - 1)(a + 1) + b^(2015)(b^2+1)(b - 1)(b + 1) + c^(2014)(c^2+1)(c - 1)(c + 1)` `⇒A=a^(2016)(a-2)(a - 1)a(a + 1)(a+2) + b^(2015)(b-2)(b - 1)a(b + 1)(b+2) + c^(2014)(c-2)(c - 1)a(c + 1)(c+2)`
tương tự trên :
`⇒(a-2)(a - 1)a(a + 1)(a+2)\vdots 5 `( vì 5 chữ số liên tiếp )
`⇒a^(2016)(a-2)(a - 1)a(a + 1)(a+2)\vdots 5`
tương tự :
`⇒b^(2015)(b-2)(b - 1)a(b + 1)(b+2)\vdots 5` và `c^(2014)(c-2)(c - 1)a(c + 1)(c+2)\vdots 5` `⇒a^(2016)(a-2)(a - 1)a(a + 1)(a+2) + b^(2015)(b-2)(b - 1)a(b + 1)(b+2) + c^(2014)(c-2)(c - 1)a(c + 1)(c+2)\vdots 5`
`⇒A\vdots 5(2)` từ` (1 ) ;(2)`
`⇒A\vdots 30`
`1/m +1/n =1/7`
`⇒(m+n)/mn=1/7`
`⇒7m +7n=mn`
`⇒7m+7n-49-mn=-49`
`⇒(m-7)(7-n)=-49`
`\begin{array}{|c|c|c|}\hline m-7 &-49 &49 &-1&1&7&-7\\\hline 7-n& 1& -1&49&-49&-7&7\\\hline n&6 & 8&-42&56&14&0\\\hline m &-42 & 56&6&8&14&0\\\hline\end{array}
vì` m;n∈N`
`⇒(m;n)∈(56;8);(8;56);(14;14)`
