Proving polynomial division examples through Mathematical Induction

Question

While proving the examples of polynomial division through mathematical induction, I am led to a curious conclusion, here is an overview:

The following statement holds true for all whole number values of 'n'.
$ \frac{x^n - y^n}{1} $ = p(x-y) -------------(1)
1. For now, let us assume 'x', 'y' to be integers
So, through the closure property of integers under multiplication and subtraction, we can state:
'p' is an integer
After substituting several whole number values of 'n' and integer values of 'x' and 'y', we find that the statement still holds true.
Thus:
Proving (1) through MATHEMATICAL INDUCTION:
To prove : $ \frac{x^(n+1) - y^(n+1)}{1} $ = a(x-y) ; or to put in other words : $ \frac{x^(n+1) - y^(n+1)}{1} $ must be represented in the form of 'a(x-y)' where 'a' must be a integer. (prove: 'a' is an integer.
Equation : $ \frac{x^n - y^n}{1} $ = p(x-y)
Assuming (1) to be true,
for n = n+1, we have: $ \frac{x^(n+1) - y^(n+1)}{1} $ = a(x-y) -----------------(2)
also we have: $ \frac{x^n - y^n}{1} $ = p(x-y) -----------------------------------(1)
Multiplying LHS and RHS of (1) with : (x+y) -
(x+y) $ \frac{x^n - y^n}{1} $ = p(x-y)(x+y)
Simplifying through index law of multiplication:
= $ \frac{x^(n+1) - y^nx + x^ny - y^(n+1) }{1} $ = $ \frac{p(x+y)(x-y)}{1} $

Rearranging the terms in the numerator of the fraction:
= $ \frac{x^(n+1)- y^(n+1) - y^nx + x^ny}{1} $ = $ \frac{p(x+y)(x-y)}{1} $
Adding LHS and RHS of (1) to : $ \frac{(y^nx)}{1}$ -
= $ \frac{x^(n+1)- y^(n+1) + x^ny}{1} $ = $ \frac{p(x+y)(x-y)+(yn^x)}{1} $
Subtracting ($ \frac{x^ny}{1}$) from LHS and RHS of (1) -
$ \frac{x^(n+1)- y^(n+1)}{1} $ = $ \frac{p(x+y)(x-y)+(y^nx)-(x^ny)}{1} $ ---------(3)
we also have another equation : $ \frac{x^(n+1) - y^(n+1)}{1} $ = a(x-y) -----------(2)
After observing (3) and (2), we can clearly state the following:
a(x-y) = $ \frac{p(x+y)(x-y)+(y^nx)-(x^ny)}{1} $ ------------------------(3)
a(x-y) = $ \frac{p(x+y)(x-y)+(y^nx)-(x^ny)}{1} $
Taking '-xy' as a common factor:

a(x-y) = $ \frac{p(x+y)(x-y)-xy(x^(n-1)-y^(n-1))}{1} $
We have tested that for all integers less than 'n' that the equation : $ \frac{x^n - y^n}{x-y} $ = (x+y), takes the form of k(x-y), where 'k' is an integer. So,

a(x-y) = $ \frac{p(x+y)(x-y)-xy(k[x-y])}{1} $
On removing the brackets and taking '(x-y)' as a common factor, we have the following:
a(x-y) = $ \frac{(x-y)[p(x+y)-xyk]}{1} $ --------------------(4)
Before going further, thoroughly read the following -
Through the closure property of integers under addition, subtraction and multiplication, we can state the following:
i). (x+y) = an integer
.....also (x-y) = an integer
ii). $ \frac{(xyk)}{1} $ = an integer
iii). p = an integer [assumed at the beginning]
Thus we can say that the expression on the RHS is an integer.
= a(integer) = integer
Now, we can state that 'a' is an integer, as an integer multiplied by an integer (x+y) must be equivalent to an integer (on the LHS).
Coming back to (4):
a(x-y) = $ \frac{(x-y)[p(x+y)-xyk]}{1} $ --------------------(4)
On Dividing both sides by the expression '(x-y)', we have the followwing:
a = p(x+y)-xyk
Substituting the value of 'a' in the (2), we have:
$ \frac{x^(n+1) - y^(n+1)}{1} $ = [p(x+y)-xyk] (x-y)
Now, we have successfully represented $ \frac{x^(n+1) - y^(n+1)}{1} $ in the form of 'a(x-y)' where 'a' is an integer.
On this wise, we can conclude that $ \frac{x^n - y^n}{1} $ is divisible by (x-y) for 'x', being integers and 'n' being a whole number; because 'a' is an integer
2. Now lets assume 'x' and 'y' to be rational numbers:
Our equation: $ \frac{x^n - y^n}{1} $ = p(x-y)
x,y = rational numbers
n = (remains)a whole number
On proving the equation for 'n+1' through mathematical induction, we get the following result:
$ \frac{x^n - y^n}{1} $ is divisible by (x-y) for 'x' and 'y' being rational numbers and 'n' being a whole number.
However, this statement is too argumentative as every non-zero rational number will divide every rational number.
Instance-1:
Let us assume we have to find whether $ \frac{21}{4} $ is divisible by $ \frac{2}{3} $
So, let the number to be multiplied with $ \frac{2}{3} $ be 'b':
($ \frac{2}{3} $)(b) = $ \frac{21}{4} $
*If 'b' is and integer, then, we can safely conclude that it $ \frac{21}{4} $ is divisible by $ \frac{2}{3}.* <br /> *But if 'b' is a rational, then it means that $ \frac{21}{4} $ isn't divisible by $ \frac{2}{3} - Hopefully, I am correct.
Solving the equation in order to obtain the value of 'b':
($ \frac{2}{3} $)(b) = $ \frac{21}{4} $
Multiplying both sides of the equation by ($ \frac{3}{2} $):
b = $ \frac{84}{8} $
Simplifying the value of 'b':
b = $ \frac{84/4}{8/4} $
b = $ \frac{21}{2} $
We see that the result obtained is a rational number, thus $ \frac{21}{4} $ isn't divisible by $ \frac{2}{3}$
Instance-2:
Now, let us find whether $ \frac{21}{12} $ is divisible by $ \frac{7}{4} $
So, let the number to be multiplied with $ \frac{7}{4} $ be 'c':
($ \frac{7}{4} $)(c) = $ \frac{21}{12} $
Solving the equation in order to obtain the value of 'c':
($ \frac{7}{4} $)(c) = $ \frac{21}{12} $
Through cross-multiplication:
c = 3
We see that the result obtained is an integer(although it is also a rational number), thus $ \frac{21}{12} $ is divisible by $ \frac{7}{4}$.
So, I think, every non-zero rational number is divisible by every other rational number. Such a division for polynomial expression where 'x', 'y' are rational numbers isn't logical.
Now, the question arises, should the equation:'$ \frac{x^n - y^n}{1} $ = p(x-y)' be considered for 'x' and 'y' being rational numbers?

Also can I safely assume that 'x' and 'y' are integers in the equation : '$ \frac{x^n - y^n}{1} $ = p(x-y)' ?

Answer 1

To show that $x-y\mid x^n-y^n$, you can factorise $x^n-y^n$ or use the Fundamental theorem of algebra. Observe that, $y=x$ is the common root of the polynomials $x-y$ and $x^n-y^n$. But you want to prove this using mathematical induction. For this, we will use the method known as "strong induction".

Let, $$p(n)=\frac{x^n-y^n}{x-y}$$

We want to prove that, $p(n)$ is a polynomial. Note that, $n=1$ is trivial. For $n=2$ you get $$\frac{x^2-y^2}{x-y}=x+y$$

Suppose that for $n=k-1$ and $n=k$, $p(k-1)=\frac{x^{k-1}-y^{k-1}}{x-y}$ and $p(k)=\frac{x^k-y^k}{x-y}$ are polynomials. Then for $n=k+1$ you have,

\begin{align}p(k+1)&=\frac{x^{k+1}-y^{k+1}}{x-y}\\ &=\frac{p(k)(x-y)(x+y)-xy(x^{k-1}-y^{k-1})}{x-y}\\ &=p(k)(x+y)-xyp(k-1).\end{align}

This completes the proof.