Re: x+y is a factor of x^(2n) - y^(2n) ?
- From: quasi <quasi@xxxxxxxx>
- Date: Mon, 07 Jan 2008 17:07:11 -0500
On Mon, 07 Jan 2008 14:29:51 EST, "Roman B. Binder"
<rbinder@xxxxxxxxxxxxxxxx> wrote:
...
but where is the proof" where I meant:
but where is the induction procedure ?
The question was how to use induction.
It suffices to prove that
(a - b) is a factor of (a^n - b^n)
for all positive integers n.
But this is easily proved by induction ...
Let f_n = a^n - b^n.
Clearly (a - b) divides f_1 and f_2.
Fix n > 2, and assume that
(a - b) | f_k for all positive integers k < n.
Next consider the expression (a + b)*f_(n-1) ...
(a + b)*f_(n - 1)
= (a + b)*(a^(n-1) - b^(n-1))
= a^n - b^n + b*a^(n-1) - a*b^(n-1)
= f_n + (a*b)*f_(n-2)
Or equivalently,
f_n = (a + b)*f_(n-1) - (a*b)*f_(n-2)
Hence by the inductive hypothesis,
since the RHS is divisible by (a - b),
it follows that f_n is divisible by (a - b),
as was to be shown.
Therefore f_n is divisible by (a - b) for all positive integers n.
Applying this to the problem at hand,
x^(2n) - y^(2n) = (x^2)^n - (y^2)^n
which is divisible by (x^2 - y^2) and hence also by (x - y).
Remarks:
The argument above deliberately avoids using The Factor Theorem, but
if The Factor Theorem is allowed to be used, that's the simplest way
to deal with this problem.
quasi
.
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