SELECTION EXAMINATION

The answer is that that maximal possible value of $c$ is $-3$.

In fact, this value is obtained if we start with the numbers $1,2,3,4,5$, and following the process: $$1,2,3,4,5\stackrel{\alpha }{\to }0,1,4,5\stackrel{\alpha }{\to }0,1,2\stackrel{\alpha }{\to }-1,0\stackrel{\beta }{\to }-3.(\ast )$$ We will prove now that no matter which are initially the numbers, we cannot have a number smaller than $-3$ during the process.

To this end, we will find an invariant under the movements. Let $w$ be the real root of the polynomial $$Px=x^3+x^2-1.$$ Then, $$w^3+w^2=1\Rightarrow w^{n+1}+w^n=w^{n-2}.(1)$$ Moreover, the polynomial $Px$ divides $Qx=x^5+x-1$, so $$Qw=0\iff w^5+w=1\iff w^{k+4}+w^k=w^{k-1}(2).$$ From (1) and (2) we observe that if we consider one extra board which will have the numbers of our board raised to the $w$ power, then the sum of the numbers in the new board is invariant under $\alpha$) and $\beta$). For example, for the sequence in (*) the extra board will have the numbers, $$w^1,w^2,w^3,w^4,w^5\stackrel{\alpha }{\to }{w}^0,w^1,w^4,w^5\stackrel{\alpha }{\to }{w}^0,w^1,w^2\stackrel{\alpha }{\to }{w}^{-1},w^0\stackrel{\beta }{\to }{w}^{-3}.$$ We observe, for example, that after the first movement the sum remains the same since $w^3+w^2=w^0$.

It is then enough to start with the minimal sum in the extra board, which is of course bigger than $$w^1+w^2+w^3+w^4+\cdots =\frac{w}{1-w}=\frac{w}{w^5}=w^{-4}.$$ So, in the initial board all the numbers are bigger than $-4$, so $c\ge -3$.