Mathematical Induction Help?

Question

Derive a divisibility conjecture for 7^(2n) - 5^(2n) (other than divisibility by 2), that is true for all n E N. prove your conjecture using mathematical induction.

Answer 1

It's always divisible by 24.

Short proof: 7^(2n) - 5^(2n) = (7^2)^n - (5^2)^n contains the

factor (7^2 - 5^2) = 24

Inductive proof: 7^(2(n+1)) - 5^(2(n+1)) =

{7^(2n) - 5^(2n)} (7^2+5^2) - 5^2*7^2(7^(2(n-1))-5^(2(n-1)))

so if true for (n-1) and n makes it true for (n+1).

Verify: 24 divides 7^2-5^2 and 7^4-5^4.



=

Answer 2

7^(2n) - 5^(2n) is divisible by 6.

Proof: Certainly this is true when n=1, since 7^2-5^2 = 49-25 = 24 = 4*6. Suppose it is true for some n∈N. Then 7^(2n) - 5^(2n) = 6*k for some natural number k. So:

7^(2(n+1)) - 5^(2(n+1))
= 49*7^(2n)-25*5^(2n)
= 24*7^(2n) + 25*7^(2n) - 25*5^(2n)
= 24*7^(2n) + 25*(7^(2n) - 5^(2n))
= 24*7^(2n) + 25*(6*k)
= 6*(4*7^(2n) + 25*k)

so 7^(2(n+1)) - 5^(2(n+1)) is divisible by 6. By induction, it follows that for all n∈N, 7^(2n) - 5^(2n) is divisible by 6. Q.E.D.

N.B. This proof doesn't have to be done by mathematical induction. If the methods of modular arithmetic are available to you, then you can use that 7^(2n) - 5^(2n) ≡ 1^(2n) - (-1)^(2n) ≡ 1-1 ≡ 0 mod 6 (which was how I first realized that the theorem was true).