Show that ## 60 ## divides ## a^{4}+59 ##

[Math100]

Homework Statement

If $gcd(a, 30)=1$, show that $60$ divides $a^{4}+59$.

Relevant Equations

None.

Proof:

Suppose $gcd(a, 30)=1$.
Then $30=2\cdot 3\cdot 5$ and $gcd(a, 2)=gcd(a, 3)=gcd(a, 5)=1$.
Applying the Fermat's theorem produces:
$a\equiv 1\pmod {2}, a^{2}\equiv 1\pmod {3}$ and $a^{4}\equiv 1\pmod {5}$.
This means $gcd(a, 4)=gcd(a, 2^{2})=1$.
Observe that
\begin{align*}
&a\equiv 1\pmod {2}\implies a^{2}\equiv 1\pmod {2}\\
&a^{2}\equiv 1\pmod {3}\implies a^{4}\equiv 1\pmod {3}.\\
\end{align*}
Now we have $2\mid (a^{2}-1)\implies a^{2}\equiv (1-2)\pmod {2}\equiv -1\pmod {2}$,
so $4\mid (a^{2}+1)(a^{2}-1)\implies 4\mid (a^{4}-1)$.
Since $60=3\cdot 4\cdot 5$ and $3, 4, 5$ are relatively prime to each other,
it follows that $60\mid (a^{4}-1)$.
Thus $a^{4}\equiv 1\pmod {60}\equiv (1-60)\pmod {60}\equiv -59\pmod {60}$.
Therefore, if $gcd(a, 30)=1$, then $60$ divides $a^{4}+59$.

 

[fresh_42]

Homework Statement:: If $gcd(a, 30)=1$, show that $60$ divides $a^{4}+59$.
Relevant Equations:: None.

Proof:

Suppose $gcd(a, 30)=1$.
Then $30=2\cdot 3\cdot 5$ and $gcd(a, 2)=gcd(a, 3)=gcd(a, 5)=1$.
Applying the Fermat's theorem produces:
$a\equiv 1\pmod {2}, a^{2}\equiv 1\pmod {3}$ and $a^{4}\equiv 1\pmod {5}$.
This means $gcd(a, 4)=gcd(a, 2^{2})=1$.
Observe that
\begin{align*}
&a\equiv 1\pmod {2}\implies a^{2}\equiv 1\pmod {2}\\
&a^{2}\equiv 1\pmod {3}\implies a^{4}\equiv 1\pmod {3}.\\
\end{align*}
Now we have $2\mid (a^{2}-1)\implies a^{2}\equiv (1-2)\pmod {2}\equiv -1\pmod {2}$,
so $4\mid (a^{2}+1)(a^{2}-1)\implies 4\mid (a^{4}-1)$.
Since $60=3\cdot 4\cdot 5$ and $3, 4, 5$ are relatively prime to each other,
it follows that $60\mid (a^{4}-1)$.
Thus $a^{4}\equiv 1\pmod {60}\equiv (1-60)\pmod {60}\equiv -59\pmod {60}$.
Therefore, if $gcd(a, 30)=1$, then $60$ divides $a^{4}+59$.

The modulo $2$ case is a bit messy.

We have $a\equiv 1\pmod{2}$ so $a^2\equiv 1\pmod{2}$. But then, $a^2+1$ and $a^2-1$ are both even, i.e. $4\,|\,(a^2-1)(a^2+1)=a^4-1$ or $a^4\equiv 1\pmod{4}.$

 

[Math100]

The modulo $2$ case is a bit messy.

We have $a\equiv 1\pmod{2}$ so $a^2\equiv 1\pmod{2}$. But then, $a^2+1$ and $a^2-1$ are both even, i.e. $4\,|\,(a^2-1)(a^2+1)=a^4-1$ or $a^4\equiv 1\pmod{4}.$

I agree, since $a$ is odd.