Question

Arătați că 7^8-3^8 este divizibil cu 10

Answer 1

$\\7^\big{8}-3^\big{8}= \\ = \Big(7^\big{4}\Big)^\big2-\Big(3^\big4\Big)^2 = \\ =\Big(7^\big4-3^\big4\Big)\cdot \Big(7^\Big4+3^\big4\Big) = \\ =\Big(\big(7^\big2\big)^\big2-\big(3^\big2\big)^\big2\Big)\cdot \Big(7^\Big4+3^\big4\Big)= \\ = \Big(7^\big2-3^\big2\Big)\cdot \Big(7^\big 2+3^\big2\Big)\cdot \Big(7^\Big4+3^\big4\Big) = \\ = \Big(7-3\Big)\Big(7+3\Big)\cdot \Big(7^\big 2+3^\big2\Big)\cdot \Big(7^\Big4+3^\big4\Big)= \\ =4\cdot 10 \cdot \Big(7^\big 2+3^\big2\Big)\cdot \Big(7^\Big4+3^\big4\Big)$

$\Rightarrow \left[4\cdot 10 \cdot \Big(7^\big 2+3^\big2\Big)\cdot \Big(7^\Big4+3^\big4\Big) \right]:10 =4 \cdot \Big(7^\big 2+3^\big2\Big)\cdot \Big(7^\Big4+3^\big4\Big) \\ \\ \\ \Rightarrow \boxed{\Big(7^\big{8}-3^\big{8}\Big) \vdots 10}$


$\\ M-am folosit de formula: \quad a^2+b^2 = (a-b)(a+b)$