Verify that ## a^{5} ## and ## a ## have the same units digit

[Math100]

Homework Statement

For any integer $a$, verify that $a^{5}$ and $a$ have the same units digit.

Relevant Equations

None.

Proof:

Let $a$ be any integer.
Applying the Fermat's theorem produces:
$a^{2}\equiv a\pmod {2}, a^{5}\equiv a\pmod {5}$.
Observe that
\begin{align*}
&a^{4}\equiv a^{2}\pmod {2}\equiv a\pmod {2}\\
&a^{5}\equiv a^{2}\pmod {2}\equiv a\pmod {2}.\\
\end{align*}
This means $a^{5}\equiv a\pmod {10}$.
Suppose $0\leq k<10$.
Then $a^{5}-k\equiv (a-k)\pmod {10}$.
Thus $a^{5}-k\equiv 0\pmod {10}\implies a-k\equiv 0\pmod {10}$.
Therefore, $a^{5}$ and $a$ have the same units digit for any integer $a$.

 

[fresh_42]

Homework Statement:: For any integer $a$, verify that $a^{5}$ and $a$ have the same units digit.
Relevant Equations:: None.

Proof:

Let $a$ be any integer.
Applying the Fermat's theorem produces:
$a^{2}\equiv a\pmod {2}, a^{5}\equiv a\pmod {5}$.
Observe that
\begin{align*}
&a^{4}\equiv a^{2}\pmod {2}\equiv a\pmod {2}\\
&a^{5}\equiv a^{2}\pmod {2}\equiv a\pmod {2}.\\
\end{align*}
This means $a^{5}\equiv a\pmod {10}$.
Suppose $0\leq k<10$.
Then $a^{5}-k\equiv (a-k)\pmod {10}$.
Thus $a^{5}-k\equiv 0\pmod {10}\implies a-k\equiv 0\pmod {10}$.
Therefore, $a^{5}$ and $a$ have the same units digit for any integer $a$.

Fine. Only one comment. As long as you consider the same modulus, it isn't necessary to repeat it. I.e. $a^5\equiv a^2\equiv a \pmod{5}$ is sufficient. All are modulo $5.$ It is different if we write $a^5\equiv a^2\pmod{5} \Longrightarrow a^5\equiv a\pmod{5}.$ The implication interrupts the calculation, the $\equiv$ sign has the same meaning as $=.$