2017 numbers - a puzzle

Question

We have $1,2,\ldots,2017$ written on a blackboard. In each step, we choose two $a\ge b$, and replace them with $a^2-b^2$. We continue to the points where there is only one number left. Which is these are true:

a) we always get an odd number at the end.

b) it is possible to obtain $0$ at the end.

c) it will always be a positive number at the end

d) we can reach any number $1,2,\ldots,2017$ in the end.

I know how to solve only a part of it. As sum is initially $2017*2018/2=2017\cdot1009$, which is odd, and in each step the sum changes by $$(a+b)-(a^2-b^2)=(a+b)(1-a+b) $$ which is always even, the sum remains odd. Hence, we end up with an even number. Hence a) is false and b) is also false. At the end, we cannot reach every number from $1$ to $2017$, as we can only reach an odd number. Hence d) is also false.

What about c)?

Answer 1

You cannot produce a negative number at any stage. Your two picked numbers are always in the order $a\geq b$, so $a^2-b^2\geq0$. You ruled out zero, so the end result will be at least one.

Answer 2

Until the phrase "$\ldots$ the sum remains odd" everything is correct.

Since the length of the list decreases by $1$ at every step the process comes to an end. But changing the odd sum $2017\cdot1009$ several times by an even number always results in an odd number, hence also the final sum and number has to be odd. It follows that a) is true and b) is false. At any move two numbers $a$, $b$ with $a\geq b$ are selected. The newly created number $a^2-b^2$ is $\geq0$. It follows that all numbers occurring in the game are $\geq0$. Since the final number is odd it has to be $\geq1$. The option d) is obviously false.