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

  • #1
960
67

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.
 

Answers and Replies

  • #3
960
67
Okay, I got ##(3^5⋅9^k-5⋅5^k)(9^k+5^k##. I'm not sure how I would proceed by induction.
 
  • #4
15,135
12,867
##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.
 
  • #5
960
67
Okay, I got it.

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

Thanks.
 

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