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Show that:
or
Take;
and
It should be|n)
Now, the proof I've found is nasty
We have
Let's define
M divides any linear combination ( with integer coeficients) of those 2.
So M divides:i.e.
Again:Thus:
And this goes on:![<br />
M\left| {\left( {\left[ {2 \cdot a^{n - 2} \cdot b + ... + 2a \cdot b^{n - 2} } \right] - 2 \cdot a^{n - 3} b \cdot \left( {a - b} \right)} \right)} \right.<br />](http://latex.codecogs.com/png.latex?<br />
M\left| {\left( {\left[ {2 \cdot a^{n - 2} \cdot b + ... + 2a \cdot b^{n - 2} } \right] - 2 \cdot a^{n - 3} b \cdot \left( {a - b} \right)} \right)} \right.<br />
)
We keep on eliminating terms, the idea is to have only 1 term in the end.
If( or the symmetric)
Whereas if n is odd:
But, since M divides a-b and (a,b)=1 thenand
thus we must have
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