<s>[INST] Proving polynomial division examples through Mathematical Induction 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)' ?
 [/INST] 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.
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