Can method of difference save the day?

From the first equation, we can see that $f(x+1)-f(x)\geq1$

$\Rightarrow f(x+2)-f(x)\geq2$ , and so on till, $f(x+5)-f(x)\geq5$

But the second equation states $f(x+5)-f(x)\leq5$ . Therefore, $f(x+5)-f(x)=5$ , or $f(x+1)-f(x)=1$ . Extrapolating a common expression, $f(n)=f(0)+n$

Therefore, $f(1000)=-496+1000=\boxed{504}$

since $f(x+5)-f(x)=5$ .

=> ${f(x+5)-f(x)}/{(5+x)-(x)}=1$ .

let $p(x,f(x)) and Q(x+5,f(x+5))$ .

so slope of pq=1

for every such points

=>f(x) must be straight line with slope=1

now $f(0)= - 496$ .

so

$f(x)= x - 496$ .

$f(1000)=504$ .

Let us start proving by contradiction that $f(x+1)-f(x)=1 \quad for\quad all\quad x\in Z \quad \quad (*)$ Assume, at the contrary, that there is an integer k such that $f(k+1)-f(k)> 1$ . Since $f(k+1)-f(k)>1$ and $f(k+2)-f(k+1)\geq1$ $f(k+3)-f(k+2)\geq 1$ $f(k+4)-f(k+3)\geq 1$ $f(k+5)-f(k+4)\geq 1,$ by adding all these inequalities and cancelling terms we would get that $f(k+5)-f(x) >5$ that would contradict the condition $f(x+5)-f(5)\leq 5$ for all integer values of $x$ .
Now we are going to prove that $(*)$ implies that $f(n)= f(0)+n \quad$ for any integer $n\geq 0$ . We can use Mathematical Induction. It is clear that this property is true when $n=0$ . Assume that $f(m)=f(0)+m$ for an integer $m\geq 0$ . Then using the latter and using (*) , $f(m+1)=f(m)+1= f(0)+m+1$ , which proves the induction thesis. Therefore $f(n)=f(0)+n$ for all non-negative integer $n$ and in particular $f(1000)=f(0)+1000=-496+1000=504.$