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QUESTION FOR FUN SOLVE FOR n: n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0 = n^n^n

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1 + n + n^2 .. n^n = n^n^n n=1 is a clear solution for n not equal to 1, we have => (n^(n+1) -1)/(n-1) = n^n^n unsure what will come next.
1^1 +1^0 = 1^1^1 2=1 which is FALSE
So, n=1 is not a solution

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Other answers:

Should I post my solution?
Oh,,Sorry about that.
Well I failed to do it doesn't mean no one will do it. Many geniuses are there on OS. I'd recommend you to wait. :)
As u say
2 minutes, I'm having a crack.
U can take your time
Before I spend the next hour looking at it to find I cant find a solution, is there definately solutions?
Apologise if it is obvious to have solutions, its been a long day of uni
But u need to PROVE it
We sure? Do you have a explanation/proof to that?
Ahh okay
I guess I have got one proof...
BY no solution, you mean no integral solution right ?
The proof that I have got is very simple
I will message u if u want
I'd like to see it. I feel there is a solution as n^n^n grows faster than a quadratic and the quadratic is smaller than the exponent around n=0
WOlfram tells me a solution, You must be meaning no integral solutions are there hmm ?
How did you enter that into wolfram? I couldnt get it to work.
Sorry, I mean integer solution of n
hmm, that is what i assumed.
Well from the question n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0 = n^n^n it is clear n>=0 and n is integer
So yea we agreed there is no solution? as the only intersection is irrational.
Now, u just need to prove it
Surely this would suffice: n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0>n^n^n \[\forall n \le 1\] and n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0
ge is supposed to read as >=
But u directly said it without any proof
Let A=n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0, and let B=n^n^n. Note: n is belonging arbitrarily to the set of integers. Then since A1, i.e. all n>=2, and A>B for all n<=1 then since there exists no integer between 1 and 2, there is no possible integer solutions to A=B
Since R.H.S is divisible by n, L.H.S must also be divisible by n. Now, all the terms except n^0 of L.H.S is divisible by n for all values of n. So,L.H.S is divisible by n only if n^0 is also divisible by n. n^0 is divisible by n only when n=1 but n=1 is not the solution. THUS, NO SOLUTION
Interesting how you look at it mroe from the abstract whilst i can see it only from the geometric side of things.
neatly done @sauravshakya

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