Thread: number sequence in a circle

Let's say we have 51 numbers located on a circle in such a way that every number is equal to the sum of its two neighboring numbers.
Prove that the only combinations of numbers is for all of them to be zero !?

It sounds simple, but I can't find a real simple proof. Please help ASAP

Hello, returnofhate!

I have half of it solved.
I'll let you solve the other half . . .

Let's say we have 51 numbers located on a circle in such a way
that every number is equal to the sum of its two neighboring numbers.
Prove that the only combination of numbers is for all of them to be zero ?

Suppose the first two numbers are and
. . and we crank out the first dozen numbers.



We see that the numbers appear in a cycle-of-6.


This means that the last terms look likt this:




The next term on the cycle would be

Since the 51 numbers lie on the circle,
. . the next term should be the first term,


And the only we can have is if


Can you finish it?

Thanks for your prompt response.

You're absolutely correct and that is a nice way.
You solved the whole thing, there is no other half, it finished as follow:

you started with an arbitrary position on the circle and showed that the starting number should be zero. This logic can be applied to any other number as the starting point and be proven to be zero. So all the numbers have to be zero (by simply starting from another position within the circle as the starting point).

BUT let's say there was not 51 elements, there was n, so can we say the same for it ?

IMO, yes we can, I was trying to prove it with induction (at least for odd numbers).
Starting from 3 points, we can simply solve the equations and show that they have to be all zero.

Then we assume there are n (odd) numbers, and the only answer is zero, now we add two more numbers, and we find the set of equations in such a way that we can use previous hypothesis to show that the original n have to be zero, and hence, the two new numbers should be zero too.

Can we prove it like that ?