The use of modulo in the mathematical induction proof of $7^{2n}-9$ being divisible by 2

Question

I tried proving that $7^{2n}-9$ is divisible by 2 in the following way:

Base case: When $n=1$,

$$7^{2}-9\equiv 0\;(\text{mod}\,2)$$

Induction step: Assume $n=k$,

$$7^{2k}-9\equiv 0\;(\text{mod}\,2)$$

We want to prove that it is true for $n=k+1$,

$$7^{2(k+1)}-9=7^{2k+2}-9=49\cdot7^{2k}-9$$

$$=49\cdot7^{2k}-9=49\cdot[0\;(\text{mod}\,2)+9]-9.$$

Since I couldn't continue, I used the following instead:

Induction step: Assume $n=k$,

$$7^{2k}-9=2a$$

For $n=k+1$,

$$7^{2(k+1)}-9=7^{2k+2}-9=49\cdot7^{2k}-9$$

$$=49\cdot7^{2k}-9=49\cdot(2a+9)-9=2(49a+216)=2b.$$

Can anyone show me how to prove this by modulo instead?

Answer 1

Yet another proof:

$7^{2n}-9 \pmod 2$ is the difference of two squares — $(7^n+9)(7^n-9)$. Both brackets are even (prove it!), so the original expression is not just even, it is divisible by $4$ (Robert Z)!

Answer 2

Following your approach, note that in the inductive step, $$7^{2(k+1)}-9=49\cdot7^{2k}-9=49\underbrace{(7^{2k}-9)}_{\equiv 0 \pmod{2}}+\underbrace{(49-1)}_{\equiv 0 \pmod{2}}\cdot9\equiv 0 \pmod{2}.$$

P.S. Instead of induction, a straightforward proof follows from the basic facts that:

1) the product of two odd numbers is odd (so $7^{2n}$ is odd);

2) the difference of two odd numbers is even (so $7^{2n}-9$ is even).

Answer 3

As an alternative simply observe that

$$7^{2n}-9\equiv (1)^{2n}-1\equiv 0 \mod 2$$