San Diego math contest problem

by DPatrick, Apr 28, 2011, 4:04 PM

Last night, I had the pleasure of attending the awards banquet for the UC San Diego High School Honors Math Contest. AoPS is a sponsor and I was helping to award the prizes.

One of the fun parts of the banquet is that they pick the best solution to each of the long-answer problems and have that student present his or her solution to the audience. However, one of the problems was sufficiently difficult on the contest such that no student completely solved it. Here it is for you to try:

Consider of function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(a^3-b^2) = f(a)^3 - f(b)^2$ holds for all integers $a,b \in \mathbb{Z}$ . What are the possible values of $f(5)$ ?

(I am told that the problem was written by UCSD math professor Dragos Oprea, who is a 4-time IMO medalist. Maybe that's why it's so hard.)

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Hey, isn't the answer the set with numbers 0, +-1, +-3, +-4???

by TheLivingShadow, Apr 30, 2011, 4:40 AM

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Answers are $0$ and $5$ . Certainly wasn't trivial.

Start by putting in $a = b = 0$ . If $f(0) = c$ , this shows that $c^3 - c^2 + c = 0$ , whose only integer root is $0$ . Hence $f(0) = 0$ .

Put in $a = 0, b = -1$ to get $f(-1) = -f(-1)^2$ , meaning $f(-1)$ is $0$ or $-1$ . Let $f(-1)$ be equal to $-e$ (so $e \in \{0,1\}$ ; note $e^2 = e$ ). Put in $a = 0, b = 1$ to get $f(1)^2 = e$ ; when we consider what $e$ can be we get $f(1) = e$ from this.

Note that if we consider the given equation as it is and also with $b \to -b$ , we find that $f(b)^2 = f(-b)^2$ . So in particular $f(b) = \pm f(-b)$ . We'll be using this a lot.

From $a = -1, b = 1$ , we get $f(-2) = -2e$ . From $a = 0, b = -2$ , we get $f(-4) = -4e$ . From $a = 2, b = 3$ we get $-e = f(2)^3 - f(3)^2 = -e = \pm 8e - f(3)^2$ . If $e = 0$ then $f(3) = 0$ . Otherwise if $e = 1$ , we get $f(3)^2$ equal to either $-7$ or $9$ , so clearly $f(3)^2 = 9$ and thus $f(3) = \pm 3e$ in all cases.

Now consider $a = 3, b = 5$ . We get $f(2) = f(3)^3 - f(5)^2$ , or $\pm 2e = \pm 27e - f(5)^2$ (the two $\pm$ are independent). If $e = 0$ then $\boxed{f(5) = 0}$ , which is one possibility achieved by $f(n) = 0$ for all $n$ . If $e = 1$ then $f(5)^2$ must be one of $-29, -25, 25, 29$ . Only $25$ is a square of an integer, so we get that $f(5)$ is one of $5, -5$ . Need some more work to tell between them; assume $e = 1$ for the rest of the proof.

Take $a = 3, b = -4$ . This gives us $f(11) = \pm 27 - 16$ . Now take $a = 5, b = 11$ , which gives $\pm 4 = f(5)^3 - (\pm 27 - 16)^2$ . By inspection, among all four possibilities here, the only way to get $f(5) = \pm 125$ (which we know) is to get $+125$ . Hence when $e = 1$ we get $\boxed{f(5) = 5}$ . This can be achieved by the solution $f(n) = n$ for all $n$ .

This post has been edited 1 time. Last edited by MellowMelon, May 3, 2011, 1:41 AM

by MellowMelon, May 2, 2011, 11:12 PM

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Art of Problem Solving Blog

San Diego math contest problem

by DPatrick, Apr 28, 2011, 4:04 PM

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