At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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
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
Actually, NO SOLUTION
But u need to PROVE it
We sure? Do you have a explanation/proof to that?
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