Hello, returnofhate!

I haveof it solved.half

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?