#4 Polynomials I

Since \(P(1)=0\), \(P(x)\) is divisible by \(x-1\). Division gives \(P(x)=(x-1)(x^2-x-6)=(x-1)(x-3)(x+2)\).

\(P(-1)=-1+2-7=-6\). The remainder is \(-6\).

\((x+4)(x-6)=x^2-2x-24\). Hence \(a=-2\), \(b=-24\).

The polynomial \(R=P-Q\) has degree at most \(2\) and three distinct roots: \(-1,0,2\). Hence \(R\equiv0\), so \(P=Q\).

For a monomial \(cx^k\), the difference of values is \(c(17^k-5^k)\), which is divisible by \(12\). The sum of such differences is also divisible by \(12\).

The value at \(x=2\) is \(8+4a-8-4a=0\). Therefore the divisibility holds for every \(a\).

Let the common root be \(r\). Then \(r^2+ar+b=0\) and \(r^2+br+a=0\). Subtracting gives \((a-b)(r-1)=0\). Since \(a\ne b\), \(r=1\).

The discriminant equals \((u+v)^2-4uv=u^2-2uv+v^2=(u-v)^2\).

If \(P\) were non-zero, it would have at most \(4\) roots. But it has at least \(5\) roots. Contradiction. Hence \(P\equiv0\).

For an integer-coefficient polynomial, \(P(8)-P(2)\) is divisible by \(6\). But the condition gives \(P(8)-P(2)=-2\), which is not divisible by \(6\). Contradiction.

The condition gives \(1000^2b+1000c+a=1000^2c+1000a+b\). Moving terms and dividing by \(999\), we get \(1001b-1000c-a=0\), so \(a=1001b-1000c\).

The value of the first trinomial at \(1\) is \(a+b+c=1002b-999c=3(334b-333c)\). It is always divisible by \(3\). The number \(2024\) is not divisible by \(3\), so this is impossible.

Let \(r\) be a common root. Then \(P(r)=0\), so \(P(P(P(r)))=P(P(0))\). By the condition this is \(0\), hence \(P(P(0))=0\).

Write \(P(x)=x^2+ux+v\). Then \(P(0)=v\), and \(P(v)=0\) gives \(v^2+uv+v=v(v+u+1)=0\).

If \(v=0\), then \(P(0)=0\). If \(v+u+1=0\), then \(P(1)=1+u+v=0\). In both cases \(P(0)P(1)=0\).

Since the product has exactly three distinct roots, the trinomials have a common root \(r\). Also \(u\ne v\), otherwise there would be only two roots.

Subtract: \(0=(r^2+ur+v)-(r^2+vr+u)=(u-v)(r-1)\). Hence \(r=1\). Substitution gives \(1+u+v=0\).

Let the other roots be \(r_1,r_2\). By Vieta, \(1+r_1=-u\), \(1+r_2=-v\). Therefore the sum of the three distinct roots is \(1+r_1+r_2=1-u-v-2=-u-v-1=0\).

Let \(P(x)=(x-a_1)(x-a_2)\), \(Q(x)=(x-b_1)(x-b_2)\). The condition says \(Q(a_1)+Q(a_2)=P(b_1)+P(b_2)\).

Expanding only the needed sums gives \((a_1-b_1)(a_1-b_2)+(a_2-b_1)(a_2-b_2)=(b_1-a_1)(b_1-a_2)+(b_2-a_1)(b_2-a_2)\).

After cancellation this is equivalent to \((a_1-a_2)^2=(b_1-b_2)^2\). For a monic quadratic, the discriminant is the square of the difference of roots. Hence the discriminants are equal.

If \(k<5\), then from any suitable polynomial \(P\) we can build another one: \(Q(x)=P(x)+(x-n_1)\cdots(x-n_k)\). It remains monic of degree \(5\), but \(Q(n_i)=P(n_i)\). Thus uniqueness is impossible.

For \(k=5\), choose \(n_i=10i\) and announce that the product of values is \(1\). The polynomial \(P(x)=\prod_{i=1}^5(x-n_i)+1\) works.

If another monic integer polynomial \(Q\) also gives product \(1\), then all \(Q(n_i)=\pm1\). But \(Q(n_i)-Q(n_j)\) is divisible by \(n_i-n_j\), while \(2\) is not divisible by \(10\) or more. Hence all values \(Q(n_i)\) are equal; since there are five of them and their product is \(1\), all equal \(1\). Then \(P-Q\) has five roots and degree at most \(4\), so \(P=Q\).

The polynomial \(h(x)=g(7-x)\) is also monic quadratic. The conditions give \(h(2)=g(5)=f(2)\) and \(h(5)=g(2)=f(5)\).

The difference \(h-f\) has degree at most \(1\), but has two roots \(2\) and \(5\). Hence \(h=f\).

If \(r\) is a root of \(f\), then \(7-r\) is a root of \(g\). The roots form two pairs with sum \(7\), so the total sum is \(14\).

The discriminants are \(D_1=p^2-4q\), \(D_2=p^2-4q-4\), \(D_3=p^2-4q-8\).

Since the first two trinomials have integer roots, \(D_1=m^2\), \(D_2=n^2\) for non-negative integers \(m,n\). Thus \(m^2-n^2=4\), so \((m-n)(m+n)=4\). The two factors have the same parity, so only \(m-n=m+n=2\) is possible, giving \(n=0\).

Then \(D_3=D_2-4=-4<0\). Therefore the third trinomial has no real roots.

By Vieta for integer roots \(r,s\): \(-b/a=r+s\), \(c/a=rs\). Hence \(a\mid b\) and \(a\mid c\).

Suppose the numbers used are \(k+1,\ldots,k+3n\). Order the leading coefficients: \(a_1<\cdots

For the trinomial with leading coefficient \(a_n\), the coefficients \(b,c\) must be positive multiples of \(a_n\), so one of them is at least \(3a_n\). But every coefficient is at most \(k+3n\). Thus \(3(k+n)\le k+3n\), hence \(k=0\). Then \(a_i=i\).

If \(i\) is even, all three coefficients in that trinomial are even. If \(i\) is odd, at least one coefficient in the triple must be even; otherwise the trinomial has odd value at every integer \(x\) and cannot have an integer root. Thus the number of even coefficients is at least \(n+2\lfloor n/2\rfloor\), which for \(n>10\) exceeds the number of even integers among \(1,\ldots,3n\). Contradiction.

The axis of the parabola passes through the midpoint of the roots, \(c=\frac{a+b}{2}\), and the vertex has coordinates \((c,-1)\). Hence \(y=\lambda(x-c)^2-1\). Substituting \(x=a\) gives \(0=\lambda\left(\frac{b-a}{2}\right)^2-1\), so \(\lambda=\frac{4}{(b-a)^2}\).

For two parabolas, common points satisfy \(\frac{4}{w_1^2}(x-c_1)^2=\frac{4}{w_2^2}(x-c_2)^2\). This is equivalent to two linear equations \(\frac{x-c_1}{w_1}=\pm\frac{x-c_2}{w_2}\). If \(w_1\ne w_2\) and \(c_1\ne c_2\), both give one distinct solution. If the width or the axis coincides, one solution merges or disappears, leaving exactly one common point.

Let \(P(a)-P(b)=N^2\). For \(P(x)=ux^2+vx+w\), \(P(a)-P(b)=(a-b)(u(a+b)+v)\).

Take \(c=a+k\), \(d=b-k\). Then \(c+d=a+b\), and \(c-d=a-b+2k\), so \(P(c)-P(d)=(a-b+2k)(u(a+b)+v)\).

If \(k=(a-b)m\), then \(P(c)-P(d)=(2m+1)N^2\). Choose \(m\) so that \(2m+1\) is a square of an odd number: \(2m+1=(2t+1)^2\). For \(t=1,2,\ldots,1001\), we get more than \(1000\) different pairs.

The polynomial \((x-a_1)(x-a_2)(x-a_3)-t\) has roots \(c_1,c_2,c_3\), so it equals \((x-c_1)(x-c_2)(x-c_3)\).

Substitute \(-x\): \((-x-a_1)(-x-a_2)(-x-a_3)-t=(-x-c_1)(-x-c_2)(-x-c_3)\).

Multiplying by \(-1\), we obtain \((x+a_1)(x+a_2)(x+a_3)+t=(x+c_1)(x+c_2)(x+c_3)\). Therefore at \(x=-a_1,-a_2,-a_3\), the left side of the equation equals \(t\). Answer: \(-a_1,-a_2,-a_3\).

The tangent to \(y=x^2\) at \(u\) is \(y=2ux-u^2\), and at \(v\) it is \(y=2vx-v^2\). Their intersection is \(x=\frac{u+v}{2}\), \(y=uv\).

Since the graphs intersect at \(u,v\), the polynomial \((p-1)x^2+qx+r\) has roots \(u,v\). Hence \((p-1)x^2+qx+r=(p-1)(x-u)(x-v)\).

Substitute \(C\) into \(y=px^2+qx+r\). Using \(q=-(p-1)(u+v)\), \(r=(p-1)uv\), we get \(uv=p\left(\frac{u+v}{2}\right)^2-(p-1)(u+v)\frac{u+v}{2}+(p-1)uv\). Simplification gives \((p-2)(u-v)^2=0\). Since \(u\ne v\), \(p=2\).

Let \(P(n)=an^2+bn+c\), \(Q(n)=un^2+vn+w\). If \(P(k)\ne P(m)\), the abscissa of the intersection of \(L_k,L_m\) is

\[x_{km}=\frac{Q(m)-Q(k)}{P(k)-P(m)}=-\frac{u(k+m)+v}{a(k+m)+b}.\]

If three lines \(L_k,L_m,L_s\) are concurrent, then \(x_{km}=x_{ks}\). Since \(m\ne s\), equality of the fractions implies \(ub-av=0\), so the expression for \(x_{km}\) does not depend on \(k,m\). Thus all pairs meet at the same abscissa \(x_0\).

Substituting \(x_0\), the values \(P(n)x_0+Q(n)\) are equal for three different \(n\). This is a quadratic polynomial in \(n\); after subtracting the common value, it has three roots, so it is identically zero. Therefore all \(L_n\) pass through one point.

By Vieta, \(r_1+r_2=-\frac ba\). Therefore

\[f(r_1+r_2)=f\left(-\frac ba\right)=a\frac{b^2}{a^2}-\frac{b^2}{a}+c=c.\]

Hence \(c=137\).