Question

Sa se afle m si n apartine N pentru care (6^m + 6^2m)(2n+1) = 6n^2 + n + 28 .

Redactare completa.Mulțumesc !!!

Answer 1

$\displaystyle Din~(6^m+6^{2m})(2n+1)=6n^2+n+28~rezulta~2n+1~|~6n^2+n+28. \\\\Avem~2n+1~|~3n(2n+1) \Leftrightarrow 2n+1~|~6n^2+3n. \\\\Din~2n+1~|~6n^2+3n~si~2n+1~|~2n+1~rezulta~ \\\\2n+1~|~(6n^2+3n)-(2n+1) \Leftrightarrow~2n+1~|~6n^2+n-1.~(*) \\\\Din~2n+1~|~6n^2+n+28~si~(*)~rezulta \\\\2n+1~|~(6n^2+n+28)-(6n^2+n-1),~deci~2n+1~|~29.$

$\displaystyle Atunci~2n+1 \in \{-29,-1,1,29 \} \Rightarrow n \in \{-15,-1,0,14\}.~Dar~n \in \mathbb{N},~deci \\\\n \in \{0,14 \}. \\ \\\bullet n=0 \Rightarrow 6^m+6^{2m}=28,~imposibil~pentru~ca~m=0~nu~convine,~iar \\\\daca ~m \ge 1,~atunci~membrul~stang~este~divizibil~cu~6. \\\\\bullet n=14 \Rightarrow (6^m+6^{2m}) \cdot 29=1218 \Rightarrow 6^m+6^{2m}=42. \\\\m=0~nu~convine. \\\\m=1~convine. \\\\m>1 \Rightarrow 6^m+6^{2m}>6^1+6^2=42.~(nu~convine) \\\\Solutie~(m,n)=(1,14).$