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

1. shubhamsrg

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.

2. sauravshakya

1^1 +1^0 = 1^1^1 2=1 which is FALSE

3. sauravshakya

So, n=1 is not a solution

4. sauravshakya

Should I post my solution?

5. shubhamsrg

6. shubhamsrg

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. :)

7. sauravshakya

As u say

8. ceb105

2 minutes, I'm having a crack.

9. sauravshakya

10. ceb105

Before I spend the next hour looking at it to find I cant find a solution, is there definately solutions?

11. ceb105

Apologise if it is obvious to have solutions, its been a long day of uni

12. sauravshakya

Actually, NO SOLUTION

13. sauravshakya

But u need to PROVE it

14. ceb105

We sure? Do you have a explanation/proof to that?

15. ceb105

Ahh okay

16. sauravshakya

I guess I have got one proof...

17. shubhamsrg

BY no solution, you mean no integral solution right ?

18. sauravshakya

The proof that I have got is very simple

19. ceb105

Shoot.

20. sauravshakya

I will message u if u want

21. ceb105

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

22. shubhamsrg

WOlfram tells me a solution, You must be meaning no integral solutions are there hmm ?

23. ceb105

How did you enter that into wolfram? I couldnt get it to work.

24. shubhamsrg
25. sauravshakya

Sorry, I mean integer solution of n

26. shubhamsrg

hmm, that is what i assumed.

27. sauravshakya

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

28. ceb105

So yea we agreed there is no solution? as the only intersection is irrational.

29. sauravshakya

Now, u just need to prove it

30. ceb105

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<n^n^n \[\forall n ge 2\] Q.E.D

31. ceb105

ge is supposed to read as >=

32. sauravshakya

But u directly said it without any proof

33. ceb105

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 A<B for all n>1, 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

34. sauravshakya

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

35. ceb105

Interesting how you look at it mroe from the abstract whilst i can see it only from the geometric side of things.

36. shubhamsrg

neatly done @sauravshakya