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?