evinda
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Hello! (Wave)
I want to show that if $m=n^{13}-n$ and $n>1$ then $30290 \mid F_m$. (Hint: Show first that $a^{13} \equiv a \mod{2730}$.)
$F_m$ is the $m$-th Fibonacci number.I have shown the hint as follows:
$2730=2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$.
Using Ferma's little theorem, we deduce that $a^{13}\equiv a \pmod{5}$, $a^{13}\equiv a \pmod{2}$, $a^{13}\equiv a \pmod{3}$, $a^{13}\equiv a \pmod{7}$ and $a^{13}\equiv a \pmod{13}$.Since $2,3,6,7,13$ are all relatively prime, we deduce that $2730 \mid a^{13}-a$.
But how can we use the fact that $a^{13} \equiv a \mod{2730}$ in order to deduce that $30290 \mid F_m$ ? (Thinking)
I want to show that if $m=n^{13}-n$ and $n>1$ then $30290 \mid F_m$. (Hint: Show first that $a^{13} \equiv a \mod{2730}$.)
$F_m$ is the $m$-th Fibonacci number.I have shown the hint as follows:
$2730=2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$.
Using Ferma's little theorem, we deduce that $a^{13}\equiv a \pmod{5}$, $a^{13}\equiv a \pmod{2}$, $a^{13}\equiv a \pmod{3}$, $a^{13}\equiv a \pmod{7}$ and $a^{13}\equiv a \pmod{13}$.Since $2,3,6,7,13$ are all relatively prime, we deduce that $2730 \mid a^{13}-a$.
But how can we use the fact that $a^{13} \equiv a \mod{2730}$ in order to deduce that $30290 \mid F_m$ ? (Thinking)