Polynomial Functional Equations

With \(y=0\), \(P(0)=0\). If the degree of \(P\) is at least \(2\), then \(P(x+y)\) has a mixed term of highest degree, absent from \(P(x)+P(y)\). Hence \(P\) is linear or zero. Thus \(P(x)=ax+b\), and \(P(0)=0\) gives \(b=0\). The answer is \(P(x)=ax\).

Let \(P\) have degree \(d\ge2\) and leading coefficient \(a\). The leading part on the left is \(a((x+y)^d+(x-y)^d)\), while on the right it is \(2ax^d\). Expanding produces a term with \(y^2x^{d-2}\) when \(d\ge2\), impossible. Hence \(d\le1\). All linear polynomials \(P(x)=ax+b\) work.

Putting \(y=0\), we get \(P(0)=0\). Comparing leading parts shows that the degree is at most \(2\). Let \(P(x)=ax^2+bx\). Substitution gives \(2ax^2+2ay^2+2bx=2ax^2+2bx+2ay^2+2by\), hence \(b=0\). From \(P(1)=2\), \(a=2\). The answer is \(P(x)=2x^2\).

The polynomial \(Q(x)=P(x)-5x\) satisfies \(Q(x+1)=Q(x)\). A periodic polynomial is constant. Hence \(Q(x)=c\), and \(P(x)=5x+c\). All such polynomials work.

Since \((x+1)^2-x^2=2x+1\), the polynomial \(Q(x)=P(x)-x^2\) is periodic with period \(1\). Therefore \(Q\) is constant. The answer is \(P(x)=x^2+c\).

The polynomial \(F(x)=\frac{x(x-1)(2x-1)}{6}\) has difference \(F(x+1)-F(x)=x^2\). Then \(Q=P-F\) is periodic with period \(1\), hence constant. The answer is \(P(x)=\frac{x(x-1)(2x-1)}{6}+c\).

The constant solutions are \(P=0\), \(P=1\). Let \(P\) be nonconstant. Then \(P(1)=1\). If \(r\ne0\) and \(P(r)=0\), then \(P(ry)=0\) for all \(y\), so \(P\equiv0\), a contradiction. Hence all roots, if any, are \(0\), and \(P(x)=ax^n\). Substitution gives \(a=a^2\), so \(a=1\). The answer is \(0\), \(1\), and \(x^n\), \(n\ge1\).

If \(P\) is constant, impossible. Let \(\deg P=d\). Then \(\deg P(P(x))=d^2\), while the right-hand side has degree \(1\). Thus \(d=1\). Let \(P(x)=ax+b\). Then \(P(P(x))=a^2x+b(a+1)\). We get \(a^2=1\), \(b(a+1)=0\). The answer is \(P(x)=x\) and \(P(x)=-x+b\), where \(b\in\mathbb R\).

A constant \(P\) is impossible. If \(\deg P=d\ge1\), then \(\deg P(P(x))=d^2\). The right-hand side has degree \(2\). We would need \(d^2=2\), impossible for an integer \(d\). Hence no such polynomial exists.

If \(d=\deg P\ge2\), then the degree of the left-hand side in \(y\) is \(d^2\), while the right-hand side has degree \(1\), impossible. Hence \(P(x)=ax+b\). Substitution gives \(a(x+ay+b)+b=ax+b+y\). Thus \(a^2=1\), \(ab=0\), so \(b=0\). The answer is \(P(x)=x\), \(P(x)=-x\).

If \(P\) is constant, then \(P=0\). Let \(d\ge1\). The degree of the left-hand side in \(y\) is \(d^2\), while the right-hand side has degree \(d\), so \(d=1\). Let \(P(x)=ax+b\), \(a\ne0\). Substitution gives \(a(x+ay+b)+b=ax+b+ay+b\). Hence \(a^2=a\), so \(a=1\); any constant works. The answer is \(P=0\) and \(P(x)=x+b\).

The constant solutions are \(0\), \(1\). Let \(P\) be nonconstant. If \(r\ne0\) is a root, then \(P(r^2)=P(r)^2=0\), and then \(P(r^{2^k})=0\) for infinitely many \(k\), impossible. Thus the only root is \(0\), and \(P(x)=ax^n\). Substitution gives \(a^2=a\), hence \(a=1\). The answer is \(0\), \(1\), and \(x^n\), \(n\ge1\).

Constant solutions satisfy \(c^3=c\), so \(c=0,1,-1\). Let \(P\) be nonconstant. A nonzero root would generate an infinite chain of roots \(r,r^3,r^9,\ldots\), so the only root is \(0\). Hence \(P(x)=ax^n\). Substitution gives \(a^3=a\), so \(a=1\) or \(a=-1\). The answer is \(0\), \(1\), \(-1\), \(x^n\), and \(-x^n\), \(n\ge1\).

Applying the condition twice gives \(P(x+2)=-P(x+1)=P(x)\). Thus \(P\) is periodic with period \(2\), so it is constant. Then \(c=-c\), hence \(c=0\). The answer is \(P\equiv0\).

For \(x^2\), \((x+1)^2-(x-1)^2=4x\). Therefore \(Q=P-x^2\) satisfies \(Q(x+1)=Q(x-1)\), i.e. \(Q(x+2)=Q(x)\). A periodic polynomial is constant. The answer is \(P(x)=x^2+c\).

The polynomial \(R(x)=P(x+1)-P(x)-x^2\) has infinitely many roots, so \(R\equiv0\). Thus the problem reduces to \(P(x+1)-P(x)=x^2\). As before, \(P(x)=\frac{x(x-1)(2x-1)}{6}+c\).

Let \(Q(x)=P(x)-x^2\). Then \(Q(x+y)=P(x+y)-(x+y)^2=P(x)+P(y)+2xy-x^2-2xy-y^2=Q(x)+Q(y)\). Hence \(Q(x)=ax\). The answer is \(P(x)=x^2+ax\).

The constant solutions are \(0\), \(1\). Let \(P\) be nonconstant. If \(r\ne0\) is a root, then \(P(r^2)=P(r)P(-r)=0\), again producing an infinite chain of roots. Thus the only root is \(0\), so \(P(x)=ax^n\). Substitution gives \(a x^{2n}=a^2(-1)^n x^{2n}\). Hence \(a=a^2(-1)^n\). Since \(a e0\), \(a=(-1)^n\). The answer is \(P(x)=x^{2k}\) and \(P(x)=-x^{2k+1}\), together with \(0\), \(1\).

If \(P\) is constant, say \(P=c\), then \(c=c^2\), so \(c=0\) or \(c=1\). Now let \(P\) be nonconstant. Then the image of \(P\) is infinite. For every number \(t\) in this image, there is \(x\) such that \(t=P(x)\), and the condition gives \(P(t)=t^2\). Hence the polynomial \(P(T)-T^2\) has infinitely many roots, so it is identically zero. Therefore \(P(x)=x^2\). Checking shows that \(0\), \(1\), and \(x^2\) all work.

Let the degree of \(P\) be \(d\), with leading coefficient \(a e0\). The leading part of the equality is \(a(x^d+y^d+(-x-y)^d)\). Put \(y=x\): we get \(a(2+(-2)^d)x^d\). If \(d\ge2\), this coefficient is nonzero: for even \(d\) it is positive, and for odd \(d\ge3\) it is \(2-2^d\). Hence \(d\le1\). Let \(P(x)=ax+b\). Substitution gives \(a(x+y-x-y)+3b=0\), so \(b=0\). The answer is \(P(x)=ax\), including \(a=0\).