orl wrote: Let $P(x)$ be a polynomial with integer coefficients. We denote $\deg(P)$ its degree which is $\geq 1.$ Let $n(P)$ be the number of all the integers $k$ for which we have $(P(k))^{2}=1.$ Prove that $n(P)- \deg(P) \leq 2.$ This problem had been proposed on AMM , i think! . Here's a simple way: We need a lemma: Lemma: Let $P(x)$ be a polynomial with integer coefficients which is not constant. Then if $P(x)$ obtains $1$ (or $-1$) as its values for at least four times then $P(x)\neq -1$ ( or $P(x)\neq 1$) for all $x$. Proof. Assume that $P(a)=P(b)=P(c)=P(d)=1$ for $a,b,c,d$ distince. Then if there's $u$ which $P(u)=-1$ then $2=P(a)-P(u)=...$ so $P(x)-P(u)-2=(x-a)(x-b)(x-c)(x-d)Q(x)$ where $Q(x)$ is a polynomial with the integer coefficients! So $-2=(u-a)(u-b)(u-c)(u-d)Q(u)$ which is impossible cause $-2$ can not presents as product of more than three distince numbers! This proved the lemma! Back to our problem: For convinet put $m=n(P)$ and $n=\deg P$. Firstly if $n\leq 2$ then $m-n\leq2$. Assume $n\geq3$. If equation $P(x)=1$ with more than three integer points (ie.. at least $4$) then equation $(P(x))^2=1$ implies $P(x)=1$ so $m\leq n$, ie... $m-n\leq 0\leq2$. The same case for equation $P(x)=-1$. So $m\leq 6$. If $n\geq4$ then $m-n\leq 6-n\leq 2$. Now assume that $n=3$. In this case if $m\leq 5$ then $m-n\leq 2$. So let us show that $m<6$. In fact if $m=6$ then $P(x)-1=0$ has three integers distince roots, and the same for $P(x)+1=0$. So $P(x)-1=k_1(x-a_1)(x-a_2)(x-a_3)$ and $P(x)+1=k_2(x-b_1)(x-b_2)(x-b_3)$ where $a_i$ distince and $b_j$ distince and all with $k_1,k_2$ are integers! Then $k_2(x-b_1)(x-b_2)(x-b_3)-k_1(x-a_1)(x-a_2)(x-a_3)=2$ for all $x$. So $k_1=k_2=k$. Finally, we have $2=k(a_i-b_1)(a_i-b_2)(a_i-b_3)$ for $i=1,2,3$ and because that $\pm1$ can not presents as products of three distince numbers so $k=\pm1$, we may assume $k=1$. Because $2=(-2)\cdot 1\cdot -1$ so $\{-2,1,-1\}=\{a_i-b_1,a_i-b_2,a_i-b_3\}$ This means $3a_i-(b_1+b_2+b_3)=-2+1-1=-2$. So we must have $3a_1=3a_2=3a_3$ which follows $a_1=a_2=a_3$, which contracts!. So $m\leq 5$ and we're done

a very silly question..... why would p(x)^2=1 be at least 4 times?? I mean...it should have maximum 2 roots(1 and -1)