# Show that a^2 + a^4 ≡ 0(mod 5) if a ≡ 2(mod 5) or a ≡ 3(mod 5) or if 5 divides a.

## Solution

### Proof.

We want to show that if a ≡ 2(mod 5) or a ≡ 3(mod 5) or if 5 divides a, then a^{2} + a^{4} ≡ 0(mod 5).

In other words, we want to show that a^{2} + a^{4} = 5k + 0, where k is an integer.

We will do this by considering each of the three cases separately.

First let’s note some theorems about modular arithmetic that will help us with our proof:

If x ≡ m(mod d) and y ≡ n(mod d), then x + y ≡ m + n(mod d).

If x ≡ m(mod d), then x

^{n}≡ m^{n}(mod d).

Case 1: a ≡ 2(mod 5)

Since a ≡ 2(mod 5), then a^{2} ≡ 2^{2} (mod 5) and a^{4} ≡ 2^{4} (mod 5).

a^{2} + a^{4} ≡ 2^{2} + 2^{4} (mod 5) ≡ 20(mod 5).

By the definition of mod, we can write a^{2} + a^{4} = 5b + 20 where b is an integer, or a^{2} + a^{4} = 5(b + 4) + 0.

Since b is an integer, b + 4 must also be an integer.

Therefore, if a ≡ 2(mod 5), then a^{2} + a^{4} ≡ 0(mod 5)

Case 2: a ≡ 3(mod 5)

Since a ≡ 3(mod 5), then a^{2} ≡ 3^{2} (mod 5) and a^{4} ≡ 3^{4} (mod 5).

a^{2} + a^{4} ≡ 3^{2} + 3^{4} (mod 5) ≡ 90(mod 5).

By the definition of mod, we can write a^{2} + a^{4} = 5c + 20 where c is an integer, or a^{2} + a^{4} = 5(c + 18) + 0.

Since c is an integer, c + 4 must also be an integer.

Therefore, if a ≡ 3(mod 5), then a^{2} + a^{4} ≡ 0(mod 5)

Case 3: 5 divides a

By the definition of divides, a = 5d, where d is an integer.

So, a^{2} + a^{4} = (5d)^{2} + (5d)^{4} = 25d^{2} + 625d^{4} = 5(5d^{2} + 125d^{4}) + 0.

Since d is an integer, 5d^{2} + 125d^{4} must also be an integer.

Therefore, if 5 divides a, then a^{2} + a^{4} ≡ 0(mod 5).