A polynomial P satisfies the following criterion:

  1. It's coefficients are integers.

  2. For all real $(a, b, c, d)$ we have $(P(a) + P(b))(P(c) + P(d)) = P(ac - bd) + P(ad + bc)$.

Determine all possible values of $P(2014)$.

My logic:

Clearly $2014$ is not a convenient value to find, so we must begin searching for possible identities of the polynomial by plugging in values, seeing what they output, and thus deducing how the polynomial behaves given certain values.

Set $(a, b, c, d) = 0$, getting

$4P(0)^2 = 2P(0)$, which, dividing by $2$, yields

$2P(0)^2 = P(0)$, and thus $P(0) = 0$, which tells us that there is no constant term and the polynomial identities we're after involve only variables and their coefficients. I write the polynomial, say degree $n$, as follows:

$Ex^n + Fx^{n-1} + Gx^{n-2} + ...$

From here on I'm kind of at a loss, because I don't know where to go or what to do. I've found the constant value, but anything else seems pretty far-fetched.


  • $\begingroup$ Please use LaTeX (MathJax) to write math. $\endgroup$ – MathMajor Dec 17 '14 at 1:53
  • $\begingroup$ I don't know how. $\endgroup$ – user164403 Dec 17 '14 at 1:54
  • 1
    $\begingroup$ I'll edit your post for you momentarily. See meta.math.stackexchange.com/questions/463/…. $\endgroup$ – MathMajor Dec 17 '14 at 1:55
  • $\begingroup$ @JimmyK4542, ah, I didn't consider that. Therefore, the constant could either be 0, or 1/2. $\endgroup$ – user164403 Dec 17 '14 at 1:57
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    $\begingroup$ Well, it looks like George took the liberty to edit the post for you. LaTeX is worth learning not only to ask questions on this site but for mathematical purposes later on. I suggest it. $\endgroup$ – MathMajor Dec 17 '14 at 2:04

Setting $(a,b,c,d) = (0,0,0,0)$ yields $4P(0)^2 = 2P(0)$, i.e. $P(0) = 0$ or $P(0) = \frac{1}{2}$.

However, since all the coefficients of $P(x)$ are integers, $P(0)$ must be an integer. Hence, $P(0) = 0$.

Then, setting $(a,b,c,d) = (x,0,y,0)$ yields $P(x)P(y) = P(xy)$ for all reals $x,y$.

Also, setting $(a,b,c,d) = (0,x,0,y)$ yields $P(x)P(y) = P(-xy)$ for all reals $x,y$.

Thus, $P(x)$ is an even polynomial. If $P(x) \not\equiv 0$, then $P(\alpha) > 0$ for some $\alpha > 0$.

Hence, $f(x) = \ln P(e^x)$ is a solution to Cauchy's functional equation $f(x+y) = f(x)+f(y)$, and is continuous at $x = \ln \alpha$. Therefore, $f(x) = nx$ for some constant $n$. This gives $P(x) = x^n$, so $n$ must be an even non-negative integer.

Setting $(a,b,c,d) = (x,1,1,1)$ yields $2(x^n+1) = (x-1)^n+(x+1)^n$. If $n > 2$ then the $x^{2}$ coefficient of the left side is $0$ but the $x^{2}$ coefficient of the right side is $2\dbinom{n}{2}$, a contradiction. If $n = 0$ we get $4 = 2$, a contradiction. Hence, the only possible value of $n$ is $n = 2$.

Clearly, $P(x) = x^2$ works. Hence the only solutions are $P(x) = 0$ and $P(x) = x^2$.

Therefore, the only possible values of $P(2014)$ are $0$ and $2014^2$.


Starting from $P(0)=0$, now set $b=d=0$ to get $P(a)P(c)=P(ac)$ and $a=c=0$ to get $P(b)P(d)=P(-bd)$ so $P$ must be even. $a=d=1, b=c=0$ gives $P(1)^2=P(1)$, so $P(1)=0,1$ $a=b=c=d=1$ gives $4P(1)^2=P(2)$ and generally $a=b=c=d$ gives $4P(a)^2=P(2a^2)$ It sure looks like the only choices are $P(x)=0$ or $P(x)=x^2$ though I haven't proven that. If we plug this in we get $$(P(a)+P(b))(P(c)+P(d))=(a^2+b^2)(c^2+d^2)\\=a^2c^2+b^2c^2+a^2d^2+b^2d^2\\=(ac-bd)^2+(ad+bc)^2\\=P(ac-bd)+P(ad+bc)$$ So $P(x)=x^2$ works


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