Science:Math Exam Resources/Courses/MATH312/December 2009/Question 06 (b)/Solution 1

The stated problem is equivalent to showing that $0\equiv a^{4n+1}-amod10$ which is equivalent to showing that 10 divides $a(a^{4n}-1)$.

To do this, we break this into cases. Since we are looking modulo 10, we only have to discuss the cases when $a\in \{0,1,2,3,4,5,6,7,8,9\}$ since all numbers reduce to one of these modulo 10. We proceed as suggested by the hints.

Case 1: $a=0$. This case clearly satisfies $a\equiv a^{4n+1}mod10$ by a simple substitution. We suppose that a is positive.

Case 2: $2\mid a$. In this case, notice that clearly 2 divides $a(a^{4n}-1)$ so it suffices to show that 5 divides $a^{4n}-1$. Notice that 5 does not divide a and so $\gcd (a,5)=1$. Thus Euler's Theorem applies giving us that

$1\equiv a^{\varphi (5)}\equiv a^{4}mod5$.

Hence

$a^{4n}-1\equiv (a^4{)}^n-1\equiv 1^n-1\equiv 0mod5$

Thus, 5 divides $a^{4n}-1$ and so 10 divides $a(a^{4n}-1)$.

Case 3: $a=5$. In this case, notice that clearly 5 divides $a(a^{4n}-1)$ so it suffices to show that 2 divides $a^{4n}-1$. This last number is even since $a^{4n}=5^{4n}$ is odd. Thus 10 divides $a(a^{4n}-1)$.

Case 4: $2\nmid a$ and $a\ne 5$. In this case, $\gcd (a,10)=1$. Thus Euler's Theorem applies giving us that

$1\equiv a^{\varphi (10)}\equiv a^{\varphi (2)\varphi (5)}\equiv a^{(1)(4)}\equiv a^{4}mod10$.

Hence

$a(a^{4n}-1)\equiv a((a^4{)}^n-1)\equiv a(1^n-1)\equiv 0mod10$

Thus 10 divides $a(a^{4n}-1)$ completing the proof.