Question

Any even power of any odd integer leaves remainder 1 when divided by 8

Answer 1

   

Hey mate!!!

Question:

Prove that any even power of any odd integer leaves remainder 1 on division by 8.

Proof:

Let the odd integer be 2x - 1, and the exponent be 2y, as it is even. So that the number becomes (2x - 1)^(2y).

Now see below. ↓↓↓↓↓

$\Rightarrow\ (2x-1)^{2y} \\ \\ \Rightarrow\ ((2x-1)^2)^y \\ \\ \Rightarrow\ (4x^2-4x+1)^y \\ \\ \Rightarrow\ (4x(x-1)+1)^y$

Here, x(x - 1) is always even, so let me take x(x - 1) as 2k.

Such that,

$\Rightarrow\ (4(2k)+1)^y \\ \\ \Rightarrow\ (8k+1)^y \\ \\ \Rightarrow\ ^yC_0(8k)^y \cdot 1^0+\ ^yC_1(8k)^{y-1} \cdot 1^1+\ ^yC_2(8k)^{y-2} \cdot 1^2+...+\ ^yC_y(8k)^0 \cdot 1^y \\ \\ \Rightarrow\ (8k)^y+\ y(8k)^{y-1}+\ ^yC_2(8k)^{y-2}+...+y(8k)+1 \\ \\ \Rightarrow\ 8[(8^{y-1} \cdot k^y)+(y \cdot 8^{y-2} \cdot k^{y-1})+(\ ^yC_2 \cdot 8^{y-3} \cdot k^{y-2})+...+yk]+1$

Here, we found in the expansion of (8k + 1)^y that the sum of all the terms except the last term 1 is divisible by 8. This means that (8k + 1)^y divided by 8 leaves remainder 1.

∴ (2x - 1)^(2y) leaves remainder 1 on division by 8.

Hence proved!!!

Plz ask me if you've any doubt on my answer.

Thank you...