Use mathematical induction to determine which of the following is divisible by 5 for all natural numbers n

Question

my work on a sheet of paper.

11 ^ ( k + 1 ) - 6 ^ ( k + 1 ) = 5a

11 ^ ( ( k + 1 ) + 1 ) - 6 ^ ( ( k + 1 ) + 1) = 5b

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11 ^ ( k + 2 ) - 6 ^ ( k + 2 )

(11 ^ k) * 11 ^ 2 - ( 6 ^ k ) * 6 ^ 2 (L)

11 ^ k - 6 ^ k = 5m

11 ^ k = 5m+6 ^ k

replacing 11 ^ k on line (L)

(5m +6 ^ k) 11 ^ 2 - 6 ^ k * 6 ^ 2

605m + 6 ^ k (11 ^ 2 - 6 ^ 2 )

from there, I cannot divide 605 by 85. I am probably missing or not understanding something

Answer 1

Prove by inducation that

11^k - 6^k is divisible by 5, for any counting integer k

k=0 : 11^0 - 6^0 = 1-1 = 0 which is divisble by 5

k=1: 11^1 - 6^1 = 11-6 = 5 which is divisible by 5

k=2: 11^2 - 6^2 = 121 - 36 = 85 which is divisible by 5

Induction Hypothesis given:

Suppose 11^n - 6^n is divisible by 5 for some positive integer n>2, n=3,4,5,6.....

11^(n+1) - 6^(n+1) = 11^n * 11 - 6^n* 6 <---- property of exponents

= 11^n * (10 + 1) - 6^n( 5+1) <--- substitution

= 11^n * 10 + 11^n - 5*6^n - 6^n <---- distributive

= 11^n *10 - 5*6^n +11^n - 6^n <--- commutative properties; swaps middle terms

= [10*11^n - 5 * 6^n] + [ 11^n - 6^n ] <--- mostly associative property; commutative

property swaps first product; then regroups

Now the first expression in brackets is divisible by 5, since FIVE (5) is factor,

in fact the GCF of the first two terms inside the first bracket pair

THe last two terms inside the second bracket pair is also divisible by 5

by virtue of the induction hypothesis.

Therefore the entire expression contains a factor of 5 which then guarantees divisibility by 5.

Therefore 11^(n+1) - 6^(n+1), from which all of this came, is also divisible by 5, by

chained equalities of the transitive property.

This concludes the proof by induction.