Cho a, b, c, d là các số nguyên thỏa mãn $a^5+b^5=4\left(c^5+d^5\right)$. CMR: a + b + c + d chia hết cho 5.

Bùi Ngọc Diễm

Theo bài ra$:$

$a^5+b^5=4\left(c^5+d\right.^5)$

$a^5+b^5+c^5+d^5=5.\left(c^5+d^5\right)\vdots5$

$\Rightarrow\left(a^5+b^5+c^5+d^5)\vdots5\right)$

Xét hiệu$:$

$a^5+b^5+c^5+d^5-\left(a+b+c+d\right)$

$=a^5-a+b^5-b+c^5-c+d^5-d$

$=a\left(a^4-1\right)+b\left(b^4-1\right)+c\left(c^4-1\right)+d\left(d^4-1\right)$

Ta có$:$

$a\left(a^4-1\right)$

$=a\left(a^2-1\right)\left(a^2+1\right)$

$=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)$

$=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a_{}+1\right)$

Vì $a,b,c,d\in Z$ và $\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)$ là $5$ số nguyên liên tiếp nên chia hết cho $5.$ $5a\left(a-1\right)\left(a+1\right)\vdots5$

$\Rightarrow\left\lbrack a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)]\vdots5\right.$

$\Rightarrow a\left(a^4-1\right)\vdots5$

Tương tự$:b\left(b^4-1\right)\vdots5;c\left(c^4-1\right)\vdots5;d\left(d^4-1\right)\vdots5$

$\Rightarrow\left\lbrack a\left(a^4-1\right)+b\left(b^4-1\right)+c\left(c^4-1\right)+d\left(d^4-1\right)\right\rbrack\vdots5$

$\Rightarrow\left\lbrack a^5+b^5+c^5+d^5-\left(a+b+c+d\right)\right\rbrack\vdots5$

Mà $\left(a^5+b^5+c^5+d^5)\vdots5\right.$ nên $\left(a+b+c+d\right)\vdots5$