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# Divisibility

Prove that

$$x^{9999} + x^{8888} + x^{7777} + x^{6666} + x^{5555} +\cdots+ x^{1111} + 1$$ is divisible by $$x^{9} + x^{8} + x^{7} +\cdots+ x + 1$$.

Note by Ankit Kumar Jain
1 year ago

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The roots of $$x^{9} + x^{8} +\ldots + 1$$ are the 10th roots of unity other than 1.

Now just show that the 10th roots of unity other than 1 are also the roots of $$x^{9999}+x^{8888}+\ldots+1$$. · 1 year ago

My solution was :

Let us call $$x^{9999} + x^{8888} + ..... + 1 = m$$ and $$x^{9} + x^{8} + ..... + 1 = n$$.

Then $$m - n = x^{9999} - x^{9} + x^{8888} - x^{8} + x^{7777} - x^{7} + ........ + x^{1111} - x$$ ............(1)

$$= x^{9}(x^{9990} - 1) + x^{8}(x^{8880} - 1) + ..... + x(x^{1110} - 1)$$

Now consider $$x^{9990} - 1 = (x^{10})^{999} - 1^{999}$$

The above expression is divisible by $$x^{10} - 1$$ because $$a^{n} - b^{n}$$ is divisible by $$(a - b)$$. Hence $$x^{9}(x^{9990} - 1)$$ is divisible by $$x^{10} - 1$$.

Similarly the following terms are divisible by $$x^{10} - 1$$ and therefore the entire expression is divisible by $$x^{10} - 1$$

Considering $$x^{10} -1 = (x-1)(x^{9} + x^{8} + ... + 1)$$.

Therefore (1) is divisible by $$x^{9} + x^{8} + ... + 1 = n$$

So (m - n) is divisible by n and hence it follows that m is divisible by n · 1 year ago

@Calvin Lin Please provide feedback for the solution... · 12 months ago

That's a nice way to see the factorization :) Staff · 12 months ago