How to prove that two exponential terms are congruent to 7?

[Eclair_de_XII]

Homework Statement

"Prove: $∀n∈ℕ,7|[3^{4n+1}-5^{2n-1}]$"

Homework Equations

The Attempt at a Solution

(1) "We take the trivial case: $n=1$, and notice that $3^5-5=238$ and $7|238$ because $7⋅(34)=238$."

(2) "Now let $n=k$ for some $1<k∈ℕ$. Then we assume that $7|[3^{4k+1}-5^{2k-1}]$. Now we must prove that $7|[3^{4(k+1)+1}-5^{2(k+1)-1}]$. We rewrite the quantity as $3^5(3^{4k})-5(5^{2k})$."

Here is where I got stuck. Basically, I was going to express everything in terms of $mod5$ and say that $3^5(3^{4k})-5(5^{2k})=2mod5$.

"Now we have $3^5(3^{4k})-5(5^{2k})=(3mod5)(1mod5)-(0mod5)(0mod5)=3mod5≠2mod5$."

I don't know how to approach the problem starting from the first sentence of (2). Can anyone help? Thanks.

 

[fresh_42]

Hint: use $(a^4x-b^2y)=(a^2x-by)(a^2+b)-a^2b(x-y)$.

 

[Eclair_de_XII]

Okay, I got $(3^5⋅9^k-5⋅5^k)(9^k+5^k$. I'm not sure how I would proceed by induction.

 

[fresh_42]

$3^5(3^{4k})−5^1(5^{2k})$

... isn't helpful. Why don't you write it as $3^4(3^{4k+1})-5^2(5^{2k-1})$ with the powers you have in your induction hypothesis? From here apply the formula I gave you and consider what you get on the right hand side.

 

[Eclair_de_XII]

Okay, I got it.

$7|(9⋅3^{4k+1}-5⋅5^{2k-1})(14)-45(3^{4k+1}-5^{2k-1})$

Thanks.