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.