Inductive Proofs for divisibility (2)

To continue our proof by induction, we now assume that $n^4-n$n4−n is divisible by $2$2 for some positive integer $n=k$n=k where $k\ge2$k≥2.

That is, we assume $\left(\editable{}\right)^4-\editable{}=\editable{}Q$()4−=Q for some integer $Q$Q.

We then aim to use this assumption to prove that $n^4-n$n4−n is divisible by $2$2 when $n=\editable{}$n=.

Using the assumption in part (c), we will now test for divisibility when $n=k+1$n=k+1.

To do so, form an expression for $\left(k+1\right)^4-\left(k+1\right)$(k+1)4−(k+1) in terms of $Q$Q.

From parts (a) and (b) we know that $n^4-n$n4−n is divisible by $2$2 when $n=\editable{}$n=.

From parts (c) and (d) we know that if $n^4-n$n4−n is divisible by $2$2 when $n=\editable{}$n= then it is divisible for $n=\editable{}$n=.

Together, these steps prove that if it works for $n=2$n=2, it also works for $n=3,4,5,6$n=3,4,5,6 etc.

Therefore, by induction, $n^4-n$n4−n is divisible by $2$2 for all positive integers $n\ge2$n≥2.

To continue our proof by induction, we now assume that $9^n-1$9n−1 is divisible by $8$8 for some positive integer $n=k$n=k.

That is, we assume $9^{\editable{}}-1=\editable{}Q$9−1=Q for some integer $Q$Q.

We then aim to use this assumption to prove that $9^n-1$9n−1 is divisible by $8$8 when $n=\editable{}$n=.

Using the assumption in part (c), we will now test for divisibility when $n=k+1$n=k+1.

To do so, form an expression for $9^{k+1}-1$9k+1−1 in factorised form, in terms of $Q$Q.

From parts (a) and (b) we know that $9^n-1$9n−1 is divisible by $8$8 when $n=\editable{}$n=.

From parts (c) and (d) we know that if $9^n-1$9n−1 is divisible by $8$8 when $n=\editable{}$n= then it is divisible for $n=\editable{}$n=.

Together, these steps prove that if it works for $n=1$n=1, it also works for $n=2,3,4,5$n=2,3,4,5 etc.

Therefore, by induction, $9^n-1$9n−1 is divisible by $8$8 for all positive integers $n$n.

To continue our proof by induction, we now assume that $5^n+7^n$5n+7n is divisible by $2$2 for some positive integer $n=k$n=k, where $k\ge2$k≥2.

That is, we assume $5^{\editable{}}+7^{\editable{}}=\editable{}Q$5+7=Q for some integer $Q$Q.

We then aim to use this assumption to prove that $5^n+7^n$5n+7n is divisible by $2$2 when $n=\editable{}$n=.

Using the assumption in part (c), we will now test for divisibility when $n=k+1$n=k+1.

To do so, form an expression for $5^{k+1}+7^{k+1}$5k+1+7k+1 in factorised form, in terms of $Q$Q.

From parts (a) and (b) we know that $5^n+7^n$5n+7n is divisible by $2$2 when $n=\editable{}$n=.

From parts (c) and (d) we know that if $5^n+7^n$5n+7n is divisible by $2$2 when $n=\editable{}$n= then it is divisible for $n=\editable{}$n=.

Together, these steps prove that if it works when $n=2$n=2, it also works when $n=3,4,5,6$n=3,4,5,6 etc.

Therefore, by induction, $5^n+7^n$5n+7n is divisible by $2$2 for all positive integers $n\ge2$n≥2.