Fijisurf wrote:
Bunuel wrote:
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RULE: for x^n-y^n:
\(x^n-y^n\) is ALWAYS divisible by \(x-y\).
\(x^n-y^n\) is divisible by \(x+y\) when n is even.
What is the reasoning behind these rules? (understanding makes it easier to remember)
Bunuel wrote:
If n is an integer >1, is 3^n-2^n divisible by 35?
(1) n is divisible by 15.
(2) n is divisible by 18.
Explanation to follow.
I just wanted to show another method to prove the above. Perfect solution is already given by everyone.
+1 for the sum.
For any function f(x), if x=a makes f(a)=0, then (x-a) is a factor of the given function.
This is something we all know. It has just been put in words by me.To illustrate that, consider the given example :
f(x) =\(x^2-2x+1\) ; for x=1, f(1) = 0. Thus, (x-1) is a factor of f(x),and as everyone would have recognised,we all know f(x) = \((x-1)^2\).
I.Function of the form f(x) = \(x^n+y^n\) will have (x+y) as a factor, if and only if for x=-y, f(-y) = 0.
Thus, replacing the value of x in the above function, we get : \((-y)^n+y^n \to\) Notice that this will be zero , only if n is odd.
\(x^n+y^n\) will always have (x+y) as a factor, if n = odd.II. Function of the form f(x) = \(x^n-y^n\). Just as above, for (x+y) to be a factor, for x = -y,f(-y) = 0
Again, by replacing f(-y) = \((-y)^n-y^n \to\) This will be zero only if n is even.
\(x^n-y^n\) will always have (x+y) as a factor, if n = even.Again, for (x-y) to be a factor, for x = y, f(y) = 0.
f(y) = \(y^n-y^n\) = 0.
\(x^n-y^n\) will always have (x-y) as a factor, if n = odd/even.