Find the largest integer that divides all integers of the ofrm 7^{2n + 1} + 1, for integers n >= 0.
The largest integer appears to be 8. It divides all integers of the form:7^(2n + 1) + 1 from n=1 to infinity.
+26367 Find the largest integer that divides all integers of the form \(7^{2n + 1} + 1\), for integers \(n \ge 0\).
My attempt:
\(\text{if }~n=0:~ 7^{2n + 1} + 1 = 7^{2*0 + 1} + 1 =7 + 1 = 8 \)
The largest integer that divides \(7^{2n + 1} + 1\), if \(n=0\) is \(8\).
The question is: \(7^{2n + 1} + 1 \equiv 0 \pmod{8} \ ?\)
\(\begin{array}{|rcll|} \hline 7^{2n + 1} + 1 &\equiv& 0 \pmod{8} \ ? \\ \hline && 7^{2n + 1} + 1 \pmod{8} \\ &\equiv& 7^{2n}*7 + 1 \pmod{8} \\ &\equiv& \left(7^2 \right)^n*7 + 1 \pmod{8} \\ &\equiv& {49}^n*7 + 1 \pmod{8} \quad |\quad 49 &\equiv& 1 \pmod{8} \\ &\equiv& {1}^n*7 + 1 \pmod{8} \\ &\equiv& 7 + 1 \pmod{8} \\ &\equiv& 8 \pmod{8} \\ &\equiv& 8-8 \pmod{8} \\ &\equiv& 0 \pmod{8} \\ \hline \end{array} \)
Find the largest integer that divides all integers of the form \(7^{2n + 1} + 1\) is 8
