I'm new to inductive proofs so I need some commentary on my proof since the book only gives a hint in the back. In "Discrete Mathematics with Applications" by Epp Third Edition in section 4.3 problem 13 states

For any integer $ n \ge 1, x^n - y^n$ is divisible by $(x - y)$ where x and y are any integers with $ x \ne y $

My Proof is as follows.

let $ Q(n) = x^n - y^n $

Then the base case is

$ Q(1) = x^1 - y^1 = 1 - 1 = 0 $

Now

$ Q(n + 1) = x^{n+1} - y^{n+1} = (x^n + y^n)(x-y)$

So now we can see $(x-y)$ is a factor and in turn divisible by $(x-y)$. I have just one hesitation. I didn't substitute from the inductive hypotheses. In every other inductive proof I've done this was a necessary step. My intuition on induction tells me that I have basically set up all of the dominoes but failed to knock down the first one (the substitution). Is this necessary for a valid proof or does this hold?

I'm new to inductive proofs so I need some commentary on my proof since the book only gives a hint in the back. In "Discrete Mathematics with Applications" by Epp Third Edition in section 4.3 problem 13 states

For any integer $ n \ge 1, x^n - y^n$ is divisible by $(x - y)$ where x and y are any integers with $ x \ne y $

My Proof is as follows.

let $ Q(n) = x^n - y^n $

Then the base case is

$ Q(1) = x^1 - y^1 $

Now

$ Q(n + 1) = x^{n+1} - y^{n+1} = (x^n + y^n)(x-y)$

So now we can see $(x-y)$ is a factor and in turn divisible by $(x-y)$. I have just one hesitation. I didn't substitute from the inductive hypotheses. In every other inductive proof I've done this was a necessary step. My intuition on induction tells me that I have basically set up all of the dominoes but failed to knock down the first one (the substitution). Is this necessary for a valid proof or does this hold?

I'm new to inductive proofs so I need some commentary on my proof since the book only gives a hint in the back. In "Discrete Mathematics with Applications" by Epp Third Edition in section 4.3 problem 13 states

For any integer $ n \ge 1, x^n - y^n$ is divisible by $(x - y)$ where x and y are any integers with $ x \ne y $

My Proof is as follows.

let $ Q(n) = x^n - y^n $

Then the base case is

$ Q(1) = x^1 - y^1 = 1 - 1 = 0 = 8(0) $

Now

$ Q(n + 1) = x^{n+1} - y^{n+1} = (x^n + y^n)(x-y)$

So now we can see $(x-y)$ is a factor and in turn divisible by $(x-y)$. I have just one hesitation. I didn't substitute from the inductive hypotheses. In every other inductive proof I've done this was a necessary step. My intuition on induction tells me that I have basically set up all of the dominoes but failed to knock down the first one (the substitution). Is this necessary for a valid proof or does this hold?

I'm new to inductive proofs so I need some commentary on my proof since the book only gives a hint in the back. In "Discrete Mathematics with Applications" by Epp Third Edition in section 4.3 problem 13 states

For any integer $ n \ge 1, x^n - y^n$ is divisible by $(x - y)$ where x and y are any integers with $ x \ne y $

My Proof is as follows.

let $ Q(n) = x^n - y^n $

Then the base case is

$ Q(1) = x^1 - y^1 = 1 - 1 = 0 $

Now

$ Q(n + 1) = x^{n+1} - y^{n+1} = (x^n + y^n)(x-y)$

So now we can see $(x-y)$ is a factor and in turn divisible by $(x-y)$. I have just one hesitation. I didn't substitute from the inductive hypotheses. In every other inductive proof I've done this was a necessary step. My intuition on induction tells me that I have basically set up all of the dominoes but failed to knock down the first one (the substitution). Is this necessary for a valid proof or does this hold?