Polynomial Vocabulary: Very Difficult Problems with Solutions

Solution:
Let [tex]E[/tex] be the sum of the coefficients of the even powers of [tex]x[/tex] (including the constant term) and [tex]O[/tex] the sum of the coefficients of the odd powers. For [tex]x=1[/tex] every power equals [tex]1[/tex], and for [tex]x=-1[/tex] the even powers equal [tex]1[/tex] and the odd powers equal [tex]-1[/tex]. Hence
[tex]P(1)=E+O , \quad P(-1)=E-O\ \Rightarrow\ E=\frac{P(1)+P(-1)}{2}[/tex]
Since the exponent [tex]2015[/tex] is odd,
[tex]P(1)=(1+1+1)^{2015}+(1+1-1)^{2015}=3^{2015}+1[/tex]
[tex]P(-1)=(1-1+1)^{2015}+(1-1-1)^{2015}=1+(-1)^{2015}=1-1=0[/tex]
Therefore
[tex]E=\frac{3^{2015}+1}{2}[/tex]

Solution:
Write [tex]P(x)=px^2+qx+r[/tex] and [tex]Q(x)=ux^3+vx^2+wx+s[/tex]. Then
[tex]P(x)(x^4+1)=px^6+qx^5+rx^4+px^2+qx+r[/tex]
[tex]Q(x)(x^3-1)=ux^6+vx^5+wx^4+(s-u)x^3-vx^2-wx-s[/tex]
Their difference
[tex](p-u)x^6+(q-v)x^5+(r-w)x^4+(u-s)x^3+(p+v)x^2+(q+w)x+(r+s)[/tex]
must be identically equal to the constant polynomial [tex]2[/tex]. Equating the coefficients of like powers of [tex]x[/tex]:
[tex]u=p , \quad v=q , \quad w=r , \quad s=u=p , \quad p+v=0 , \quad q+w=0 , \quad r+s=2[/tex]
Hence [tex]q=-p[/tex], [tex]r=-q=p[/tex] and [tex]r+s=2p=2[/tex], so [tex]p=1[/tex], [tex]q=-1[/tex], [tex]r=1[/tex] (and then [tex]Q(x)=x^3-x^2+x+1[/tex]). Therefore
[tex]P(x)=x^2-x+1[/tex]
Check:
[tex](x^2-x+1)(x^4+1)-(x^3-x^2+x+1)(x^3-1)=[/tex]
[tex]=\left(x^6-x^5+x^4+x^2-x+1\right)-\left(x^6-x^5+x^4+x^2-x-1\right)=2[/tex]

Solution:
The denominators are non-zero because [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] are distinct. Each of the three terms is a constant times a product of two linear factors, so [tex]F(x)[/tex] is a polynomial of degree at most [tex]2[/tex], and so is [tex]G(x)=F(x)-x^2[/tex]. Write [tex]G(x)=px^2+qx+r[/tex]; we must show that [tex]p=q=r=0[/tex].
At [tex]x=a[/tex] the second and third terms vanish (they contain the factor [tex]x-a[/tex]), and the first term becomes
[tex]\frac{a^2(a-b)(a-c)}{(a-b)(a-c)}=a^2[/tex]
So [tex]F(a)=a^2[/tex], and in the same way [tex]F(b)=b^2[/tex] and [tex]F(c)=c^2[/tex]. Hence [tex]G(a)=G(b)=G(c)=0[/tex], i.e.
[tex]pa^2+qa+r=0 , \quad pb^2+qb+r=0 , \quad pc^2+qc+r=0[/tex]
Subtracting the second equation from the first gives [tex]p(a^2-b^2)+q(a-b)=0[/tex], and dividing by [tex]a-b\ne 0[/tex] we get [tex]p(a+b)+q=0[/tex]. In the same way the first and third equations give [tex]p(a+c)+q=0[/tex]. Subtracting these two equalities:
[tex]p(b-c)=0[/tex]
Since [tex]b\ne c[/tex], [tex]p=0[/tex]. Then [tex]q=-p(a+b)=0[/tex], and from [tex]pa^2+qa+r=0[/tex] also [tex]r=0[/tex].
So [tex]G[/tex] is the zero polynomial, i.e. all coefficients of [tex]F(x)[/tex] coincide with those of [tex]x^2[/tex], and [tex]F(x)=x^2[/tex].