49th Mathematical Olympiad in Ukraine

Take $t=5k$ and $z=7k$, then we want the equality $x^2+y^2=k^2$ to hold. Consider two series of solutions: $x=n^2-1$, $y=2n$, $k=n^2+1$ and $x=3n^2+4n+1$, $y=2n(2n+1)$, $k=5n^2+4n+1$. It is evident that in both cases $(x,y,z,t)=1$ (here and on $(x,y,z,t)$ denotes $\text{GCD}\{x,y,z,t\}$) for even $n$. Consider the first series and denote $Q=2P^2(t)+2P(xy+yz+zx)-P^2(x+y+z)$ and $P(x)=a_m{x}^m+a_{m-1}{x}^{m-1}+\cdots +a_{1}x+a_0$, $a_m\ne 0$. Then polynomial $Q_1(n)=2P^2(5(n^2+1))+2P(7n^4+16n^3-12n-7)-P^2(8n^2+2n+6)$ equals to zero for every $n=2s$ with $s\in \mathbb{N}$, whence $Q_1(n)=0$. Consider the coefficient at $x^{4m}$ in $Q_1(x)$: on the one hand it is equal to $2\cdot 5^{2m}{a}_m^2+2\cdot 7^m{a}_m-8^{2m}{a}_m^2$ and on the other $0$, therefore $2\cdot 5^{2m}{a}_m^2+2\cdot 7^m{a}_m-8^{2m}{a}_m^2=0$. Analogously we can obtain that $Q_2(n)=2P^2(5(5n^2+4n+1))+2P(257n^4+T_2(n))-P^2(42n^2+34n+8)$, with $\mathrm{deg}(T_2)\le 3$, is equal to zero and thus the coefficient at $x^{4m}$ in this polynomial also equals zero: $2\cdot 25^{2m}{a}_m^2+2\cdot 257^m{a}_m-42^{2m}{a}_m^2=0$. So, $a_m=\frac{2\cdot 7^m}{64^m-2\cdot 25^m}=\frac{2\cdot 257^m}{42^{2m}-2\cdot 25^{2m}}$, whence $2\cdot 7^m\cdot (42^{2m}-2\cdot 25^{2m})=(64^m-2\cdot 25^m)\cdot 2\cdot 257^m$. For $m\ge 2$ we have: $(64^m-2\cdot 25^m)\cdot 2\cdot 257^m>(64^m-2\cdot 25^m)\cdot 2\cdot 252^m$, and so $42^{2m}-2\cdot 25^{2m}>36^m(64^m-2\cdot 25^{2m})$. $7^{2m}>7^{2m}-\frac{2\cdot 25^{2m}}{36^m}>64^m-2\cdot 25^m=(7^2+15{)}^m-2\cdot 25^m>7^{2m}+m\cdot 7^{2m-2}\cdot 15-2\cdot 25^m\Rightarrow 2\cdot 25^m>m\cdot 49^{m-1}\cdot 15\ge 2\cdot 49^{m-1}\cdot 15\Rightarrow \frac{49^{m-1}}{15}\ge \frac{49}{25}$, which is evidently false. Thus, $m\le 1$. For $m=1$, we have $P(x)=a_{1}x+a_0$ and $a_1=1$, i.e. $P(x)=x+a_0$. Substitute $P(x)$ into the initial equality and take $(x,y,z,t)$ from the first series: $(n^2-1,2n,7(n^2+1),5(n^2+1))$. After some easy transformations we will have: $4a_0(5n^2+5)+2a_0^2+2a_0-2a_0(8n^2+2n+6)-a_0^2=0$, which implies that $a_0=0$ or that $a_0$ depends on $n$, which is impossible. So for $m=1$ we have solution $P(x)=x$. Now if $n=0$ then $a_0^2+2a_0=0$, and we obtain two more answers: $P(x)=0$ and $P(x)=-2$. An easy check shows that these solutions meet all the requirements.

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Alternative solution.

Consider the equation $x^2+y^2+z^2=2t^2$ and make use of the method of Diophantine secants: ${\left(\frac{x}{t}\right)}^2+{\left(\frac{y}{t}\right)}^2+{\left(\frac{z}{t}\right)}^2=2$, i.e. ${\alpha }^2+{\beta }^2+{\gamma }^2=2$, where $\alpha ,\beta ,\gamma \in {\mathbb{Q}}^{+}$. Evidently, $\alpha =0$, $\beta =\gamma =1$ is a solution of the last equation. Now take $\beta =-k_1\alpha +1$, $\gamma =-k_2\alpha +1$, with $k_1,k_2\in \mathbb{Q}$. Then ${\alpha }^2+(1-k_1\alpha {)}^2+(1-k_2\alpha {)}^2=2\iff \alpha (k_1^2\alpha +k_2^2\alpha -2k_1-2k_2+\alpha )=0$. If $\alpha \ne 0$, we have $\alpha =\frac{2(k_1+k_2)}{1+k_1^2+k_2^2}\Rightarrow \beta =\frac{1+k_2^2-k_1^2-2k_1{k}_2}{1+k_1^2+k_2^2}$, $\gamma =\frac{1+k_1^2-k_2^2-2k_1{k}_2}{1+k_1^2+k_2^2}$. Take $k_1=\frac{n}{p}$, $k_2=\frac{m}{p}$, $n,m,p\in \mathbb{N}$. Then $\alpha =\frac{2(m+n)p}{p^2+n^2+m^2}$, $\beta =\frac{p^2+m^2-n^2-2mn}{p^2+n^2+m^2}$, $\gamma =\frac{p^2+n^2-m^2-2mn}{p^2+n^2+m^2}$. Thus, we see that $x=2(m+n)p$, $y=p^2+m^2-n^2-2mn$, $z=p^2+n^2-m^2-2mn$, $t=p^2+m^2+n^2$, with $n,m,p\in \mathbb{N}$ are the solutions of the equation $x^2+y^2+z^2=2t^2$. First consider $m=2$, $n=1$. Then $x_1=6p$, $y_1=p^2-1$, $z_1=p^2-7$, $t_1=p^2+5$. It is evident that for $p\ne 6$ we have $(x_1,y_1,z_1,t_1)=1$. Then substituting these $(x_1,y_1,z_1,t_1)$ into the problem statement we obtain that $Q=2P^2(t)+2P(xy+yz+zx)-P^2(x+y+z)=0$, which means that $Q_1(p)=2P^2(p^2+5)+2P(p^4+12p^3-8p^2-48p+7)-P^2(2p^2+6p-8)=0$. From the problem conditions it follows that all numbers $p$ of the form $p=6k$ should be the roots of $Q_1$, thus $Q_1\equiv 0$. Let $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$, with $a_n\ne 0$. Then $Q_1(p)=2(a_n(p^2+5{)}^n+\cdots +a_0{)}^2+2(a_n(p^4+12p^3-8p^2-48p+7{)}^n+\cdots +a_0{)}^2-(a_n(2p^2+6p-8{)}^n+\cdots +a_0)=(2a_n^2+2a_n-2^{2n}{a}_n^2)p^{4n}+R_1(p)$, where $\mathrm{deg}{R}_1\le 4n-1$. Since $Q_1(p)\equiv 0$ and $a_n\ne 0$ we have that $$2a_n+2-2^{2n}{a}_n=0.(1)$$ Consider $p=2m$, $n=1$. Then $x_2=4m(m+1)$, $y_2=5m^2-2m-1$, $z_2=3m^2-2m+1$, $t_2=5m^2+1$. For $m$ large enough we have $t_2>y_2>x_2>z_2$, and so these numbers are distinct. If $m$ is even, $(x_2,z_2)=(4m(m+1),3m^2-2m+1)=(4(m+1),3m^2-2m+1)=(m+1,3m^2-2m+1)=(m+1,5m-1)=(m+1,6)$, thus for $m\ne 6$ $(x_2,y_2,z_2,t_2)=1$. If we follow the same lines as for the first series $(x,y,z,t)$, we will obtain that $Q_2(m)=2P^2(5m^2+1)+2P(47m^4+T(m))-P^2(12m^2)=0$, where $\mathrm{deg}T\le 3$. Let us compute the coefficient at $m^{4n}$ in $Q_2(m)$ - it's equal to $2a_n^2\cdot 5^{2n}+2a_n\cdot 47^n-12^{2n}{a}_n^2=0$. Since $a_n\ne 0$ we have that: $$2a_n\cdot 5^{2n}+2\cdot 47^n-12^{2n}{a}_n=0.(2)$$ Now from (1) and (2) we obtain that $a_n=\frac{1}{2^{2n-1}}$ and $(2^{2n-1}-1)\cdot 2\cdot 47^n=12^{2n}-2\cdot 5^{2n}$, whence $2\cdot 5^{2n}+(4\cdot 47{)}^n=144^n+2\cdot 47^n$. It's easy to see that the last equality is impossible for arbitrary natural $n\ge 2$, since $2\cdot 5^{2n}+(4\cdot 47{)}^n>188^n=(144+44{)}^n\ge 144^n+44\cdot 144^{n-1}>144^n+2\cdot 47^n$. So $\mathrm{deg}P\le 1$. First assume that $\mathrm{deg}P=1$. Then the equality $a_n=\frac{1}{2^{2n-1}}$ implies that $a_1=1$, i.e. that $P(x)=x+a_0$. Substitute this polynomial into the initial equality: $$\begin{gathered} 2P^2(t)+2P(xy+yz+zx)=P^2(x+y+z)\Rightarrow \\ 2(t^2+2a_{0}t+a_0^2)+2(xy+yz+zx)+2a_0=(x+y+z{)}^2+2(x+y+z)a_0+a_0^2\iff \\ 4a_{0}t+a_0^2+2a_0=2(x+y+z)a_0. \end{gathered}$$ If $a_0=0$ then $P(x)=x$, otherwise $4t+a_0+2=2(x+y+z)$. Substitute here the known series $x_1=6p$, $y_1=p^2-1$, $z_1=p^2-7$, $t_1=p^2+5$ with $p=6$. Then $4(p^2+5)+a_0+2=2(2p^2+6p-8)$, and we get that the constant $a_0$ depends on $p$, which is impossible. If $\mathrm{deg}P=0$ then $P=a_0=\text{const}$. From the problem conditions we obtain that $2a_0^2+2a_0=a_0^2\Rightarrow a_0=-2$ also $a_0=0$, which means that $P(x)=0$ or $P(x)=-2$. An easy check shows that all three polynomials obtained meet all the requirements.