By "nonzero" polynomial do you mean a polynomial which never equals zero? If so... $d(x) = \frac{(p(x))^2 - 1}{(q(x))^2} = x^2 + ax + b$. $deg(p(x)) - deg(q(x)) = 2$ which yields: $deg(p(x)) = 2, \ deg(q(x)) = 1$ or $deg(p(x)) = 1, \ deg(q(x)) = 0$. We must discard the first possibility since q(x) would have a zero. Thus, $p(x) = ax+b, \ p(x) = a^2x^2 + 2abx + b^2, \ q(x) = a \ne 0$ $d(x) = \frac{a^2x^2 + 2abx + b^2-1}{a^2} = x^2 + \frac{2b}{a}x + \frac{b^2-1}{a^2}$ The third term yields $b^2 = na^2+1$ which I daresay has only the solutions $b = 1, \ -1$ corresponding to n = 0 (please correct me if I err). It follows from the second term that $a = \pm 1, \ \pm 2$. Thus, $d(x) = x^2 \pm x$ and $d(x) = x^2 \pm 2x$. If "nonzero" means that q is nontrivial, then we have some additional work to do.

I think he mean that non of coefficients of $q(x)$ is zero.

mathisfun1 wrote: By "nonzero" polynomial do you mean a polynomial which never equals zero? If so... $d(x) = \frac{(p(x))^{2}-1}{(q(x))^{2}}= x^{2}+ax+b$. $deg(p(x))-deg(q(x)) = 2$ which yields: $deg(p(x)) = 2, \ deg(q(x)) = 1$ or $deg(p(x)) = 1, \ deg(q(x)) = 0$. Why?

mathisfun1 wrote: By "nonzero" polynomial do you mean a polynomial which never equals zero? If so... $ d(x) = \frac {(p(x))^2 - 1}{(q(x))^2} = x^2 + ax + b$. $ deg(p(x)) - deg(q(x)) = 2$ which yields: $ deg(p(x)) = 2, \ deg(q(x)) = 1$ or $ deg(p(x)) = 1, \ deg(q(x)) = 0$. We must discard the first possibility since q(x) would have a zero. Thus, $ p(x) = ax + b, \ p(x) = a^2x^2 + 2abx + b^2, \ q(x) = a \ne 0$ $ d(x) = \frac {a^2x^2 + 2abx + b^2 - 1}{a^2} = x^2 + \frac {2b}{a}x + \frac {b^2 - 1}{a^2}$ The third term yields $ b^2 = na^2 + 1$ which I daresay has only the solutions $ b = 1, \ - 1$ corresponding to n = 0 (please correct me if I err). It follows from the second term that $ a = \pm 1, \ \pm 2$. Thus, $ d(x) = x^2 \pm x$ and $ d(x) = x^2 \pm 2x$. If "nonzero" means that q is nontrivial, then we have some additional work to do. Sorry,but I don't understand a few things. We get $ degP-degQ=1$,why did you just consider the case $ degP=2,degQ=1$,what about the other cases?Like $ degP=3,degQ=2$?Thanks.

That is my solution. Maybe too long If a is odd then $4b=a^{2}-1$ and if a is even then $b=\frac{a^{2}}{4}+k$ with $k\mid 2$ First: if $a=2r$ $d(x)= (a+r)^{2}+k$ here $k=b-r^{2}$ Let $f(x)=P(x+r), g(x)=Q(x+r)$ then we have: $(P(x))^{2}-(x^{2}+k)(Q(x))^{2})=1$ $R(x)=P(x)-xQ(x)$ we have $R(x).(2xQ(x)+R(x))=1+k(Q(x))^{2}$(*) Easy to see that: If $deg(P)=2$ then only root for (*) in $Q_{[x]}$ is $Q(x)=\frac{2}{k}.x, P(x)=1+2/k.x^{2}$ we have root in $Q_{[x]}$ with deg(P)=1 if and only if $k=-1/m^{2}$ and $m\in Q^{+}$ and only root in that case is; $P(x)=mx.Q(x)=m$ Let $P_{1},Q_{1}$ is root of (*) in $Q_{[x]}$ and $deg P$ is min and $Q_{1}$ not equal 0 If $deg(P_{1})=n\geq 3$ $R_{1}(x).(2xQ_{1}(x)+R_{1}(x))=1+k(Q_{1}(x))^{2}$(**) Let $Q^{*}(x)=\frac{2xR(x)}{k}-Q(x)$ with (**)then $deq Q^{*}=n-2>0$ and $R(x), Q^{*}$ is a root of (**) hence $Q^{*}, P^{*}$ is a root of (*) $P^{*}= xQ^{*}+R(x)$ And $deg(P^{*})=n-1$ .That is not true. Then $deqP_{1} \leq 2$ If k not equal $-m^{2}$ , $m \in Z^{+}$ Hence $Q_{1}(x)=\frac{2}{k}.x, P_{1}(x)=1+2/k.x^{2}$ is minnimal root Introduce with deq P we have that: All roots for (*) in $Q_{[x]}$ $P_{n}(x)=2P_{0}P_{n-1}(x)-P_{n-2}(x), Q_{n}(x)=2P_{0}Q_{n-1}(x)-Q_{n-2}(x)$ and $P_{0}=1, Q_{0}=0,$ $Q_{1}(x)=\frac{2}{k}.x, P_{1}(x)=1+2/k.x^{2}$ and Easy too see max degree coefficient of $P_n$ if $\frac{2^{2n-1}}{k^{n}}$ then (*) have root in $Z_{[x]}$ iff $k\mid 2$ if $k=-m^{2}$ similar we have All roots for (*) in $Q_{[x]}$ $P_{n}(x)=2P_{0}P_{n-1}(x)-P_{n-2}(x), Q_{n}(x)=2P_{0}Q_{n-1}(x)-Q_{n-2}(x)$ and $P_{0}=1, Q_{0}=0,$ $P(x)=\frac{1}{m}x, Q(x)=\frac{1}{m}$ max degree coefficient of $P_n$ if $\frac{2^{n-1}}{m^{n}}$ then (*) have root in $Z_{[x]}$ iff $m\mid 2$ and $\mid m \mid <2$ Hence $m=1$ then $k=-1$ End of proof with a is even. Case of a is odd similar, note more $k=4b-a^{2}=4r-1$ If a is odd then $4b=a^{2}-1$ and if a is even then $b=\frac{a^{2}}{4}+k$ with $k\mid 2$ Easy too see a solution: $Q(x)=\frac{2}{k}.x, P(x)=1+2/k.x^{2}$ if a even $P(x)=mx+n$ with a odd QED