Induction/6^(n+2)+7^(2n+1) divisible by 43/Exercise/Solution

Initial case. For ${}n=0$ we have

${}6^{2}+7=43\,,$

which is a multiple of ${}43$. Induction step. Suppose that the statement is proved for ${}n$ and consider the term for ${}n+1$. It is

${}{\begin{aligned}6^{n+1+2}+7^{2(n+1)+1}&=6\cdot 6^{n+2}+7^{2}\cdot 7^{2n+1}\\&=6\cdot 6^{n+2}+(6+43)7^{2n+1}\\&=6{\left(6^{n+2}+7^{2n+1}\right)}+43\cdot 7^{2n+1}\\&=6\cdot 43\cdot s+43\cdot 7^{2n+1}\\&=43\cdot {\left(6\cdot s+7^{2n+1}\right)},\end{aligned}}$

where in the penultimate step we have used the induction hypothesis (the property, that ${}6^{n+2}+7^{2n+1}$ is a multiple of ${}43$). Therefore this number is also a multiple of ${}43$.

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