**Problem**: Find all functions $ f(x) $ that satisfy $ f(x)f(y)-f(xy) = x+y $ for all reals $ x,y $.

**Solution**: Let's start by plugging in something easy, like $ x = y = 1 $. We get

$ [f(1)]^2-f(1) = 2 $

$ [f(1)-2][f(1)+1] = 0 $

so $ f(1) = 2, -1 $. Hmm, make it a little more complicated then and try just $ y = 1 $. Then

$ f(x)f(1)-f(x) = x+1 $

$ f(x)[f(1)-1] = x+1 \Rightarrow f(x) = \frac{x+1}{f(1)-1} $.

But wait, we already know $ f(1) = 2, -1 $ so we can just plug this in. Thus we have

$ f(x) = x+1 $ or $ f(x) = -\frac{x+1}{2} $.

We check by plugging each in:

$ f(x)f(y)-f(xy) = (x+1)(y+1)-(xy+1) = x+y $ (this works)

$ f(x)f(y)-f(xy) = \left(-\frac{x+1}{2}\right) \left(-\frac{y+1}{2}\right) - \left(-\frac{xy+1}{2} \right) \neq x+y $ (this doesn't).

That means only $ f(x) = x+1 $ works. QED.

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Comment: Functional equations are pretty strange as far as problems go. Basically you go around plugging random things in until you find something useful. Then you work it all out and you usually get an interesting result.

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Practice Problem: (360 Mathematical Contests - 1.1.49) Find all polynomials $ P(x) $ with integral coefficients such that

$ P(P^{\prime}(x)) = P^{\prime}(P(x)) $

for all real numbers $ x $.

Just a note to commentors: Don't put links in your comments. Spammers have been posting on my blog lately, so if you put a link it most likely will get spam-blocked.

ReplyDeleteP(x) = x

ReplyDeleteanalyze degree and leading coefficient to see deg P = 1, then its EZ

OK, I think this is the best way to do this; if you want to post a comment, just register quickly and it should work fine.

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ReplyDeletedo you happen to be able to know all the passwords?? just curious

No, I don't know the passwords... then I could be an identity thief and take over the world or something.

ReplyDelete