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**alter ego****Member**- Registered: 2012-03-30
- Posts: 27

Q2.

To make it simpler to type I'll replace 7^7 with A

The problem becomes (A^7+1)/(A+1)

= (A+1)^7/(A+1) - (7A^6 + 21A^5 + 35A^4 + 35A^3 + 21A^2 + 7A^1)/(A+1)

=(A+1)^6 - 7A^5 - 14A^4 - 21A^3 - 14A^2 -7A^1

= A^6 - A^5 + A^4 - A^3 + A^2 - A + 1

I thought I had this factorised but, in typing, I've spotted a mistake so I'll stop for now and sleep on it.

Later edit

I have used WolframAlpha and there is factoisation but it is not simple. To check the algebra I substituted A = 2 and then subsequently A = 7, at the start and at the end of this simplification. I got exactly the same value at each end of the algebra. The chances of this happening when there is, in fact, an error, is very small.

It is of course possible that this changes when A is replaced by 7^7, so I'll keep thinking.

5 minutes later edit.

Replacing A as suggested gives 7 Prime factors, the lowest of which is 197. So, many numerical solutions are possible. The question wording IMPLIES that this is not what is expected but rather a non evaluated solution that 'drops' out of the working somehow. Still thinking.

Bob

*Last edited by alter ego (2019-01-05 23:15:59)*

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**alter ego****Member**- Registered: 2012-03-30
- Posts: 27

Strange result. Not exactly 'fun' but it may make you laugh.

My wife took one look at my algebra and said "It looks like a sequence."

"Of course", I thought, "a geometric series!"

So I set to work:

Let S = a^6 - a^5 + a^4 - a^3 + a^2 - a + 1

then aS = a^7 - a^6 + a^5 - a^4 + a^3 - a^2 + s

adding : S + aS = a^7 +1

So S =(a^7 + 1)/(a+1)

!!!!!

And to think I did all that complicated binomial expanding.

Ho hum. Back to the drawing board.

Bob

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