Question

Show that $3^{4n+2}+5^{2n+1}$ is a multiple of $14$.

Answer 1

$f\left(n\right)=3^{4n+2}+5^{2n+1}f\left(1\right)=3^{4+2}+5^{2+1}=729+125=854=14\times 61$

which is a multiple of $14$

$f(n+1)=3^{4n+6}+5^{2n+3}f(n+1)-25f\left(n\right)=3^{4n+6}+5^{2n+3}-25(3^{4n+2}-5^{2n+1})f(n+1)-f\left(n\right)=3^{4n+6}-{25.3}^{4n+2}+5^{2n+3}-{25.5}^{2n+1}f(n+1)-f\left(n\right)=3^{4n+2}(81-25)+5^{2n+3}-5^{2n+3}f(n+1)-f\left(n\right)=56.3^{4n+2}$

which is also a multiple of $14$

Hence proved.