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Let f,g be polynomials with rational coefficients. Suppose f(Q) = g(Q) (i.e. the sets of values of f and g on the rationals are the same). Then f(x) = g(ax + b) for some rational constants a,b.
Remark. It generalizes to number fields with real embeddings.
Note that the converse is trivial.
Solution
The idea is that (mod the full details) there's a fixed discrete subgroup L of Q such that for any integer x, the rational solutions, viewed in y, of g(y) = f(x) are all in L. To extract the behavior of y as x varies, and compare the coefficients at the same time, i.e. the growth rates, you can consider expressions of the form g(y') - g(y) = f(x+1) - f(x) where y' is the corresponding rational number such that g(y') = f(x+1). By mean value theorem (don't quite need this, since f,g are polynomials so one can potentially do all the computations by algebra) the expression extracts out y' - y in terms of f'(x) and g'(y) roughly, and this gives a good comparison of y as x varies. This is the idea, and I do some computations to show that this works.
Also, simple valuation calculations (plug in this, plug in that rational number, compare the p-adic valuations) don't really seem to get anywhere, as I give plenty of counter examples and point out the problems to the claims of one of the posters below.
 
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On September 11 2010 05:34 Assault_1 wrote:
so whats the question?
Proving it.
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Would a non-math major/non-math enthusiast be able to solve this?
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the sets f(Q) and g(Q) are not counting multiplicity right?
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On September 11 2010 05:52 Piege wrote:
Would a non-math major/non-math enthusiast be able to solve this?
I don't know the solution yet but I suspect the answer to your question is no.
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On September 11 2010 05:53 category wrote:
the sets f(Q) and g(Q) are not counting multiplicity right?
They're just sets, not really any structures on them (for ex. ramifications). For example for the functio f(x) = x^2 the set f(Q) is just the set of all squares of rational numbers. We don't care that f(x) = 0 ramifies (and elsewhere unramifies). It's just a naive set.
I wonder if that's what you meant. So yes, I think we're not counting multiplicities here, whatever you mean by that in this context. f(Q) = {y : y = f(a) for some rational a}.
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lol. I don't know about ramifications. i just meant, eg in the f(x^2) case, does it matter that 1 is obtained twice? but it sounds like no.
i guess for the functions f=x^2 and g=x^4, f(Q) will be larger than g(Q), because some rationals are the square of another rational but are not a fourth power?
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On September 11 2010 05:59 category wrote:
lol. I don't know about ramifications. i just meant, eg in the f(x^2) case, does it matter that 1 is obtained twice? but it sounds like no.
Right, just count it once. So for example {1,1} = {1}.
i guess for the functions f=x^2 and g=x^4, f(Q) will be larger than g(Q), because some rationals are the square of another rational but are not a fourth power?
That's right.
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Then f(x) = g(ax + b) for some rational constants a,b.
whats the domain of x, rational or real?
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On September 11 2010 06:04 Assault_1 wrote:whats the domain of x, rational or real?
any field containing the rationals. It really doesn't matter, since the condition is only imposed on f(Q) and g(Q). You don't need to "see" bigger fields containing the rationals. Any such field is infinite anyway, so if the condition f(x) = g(ax + b) holds for evaluating x on all rationals then it's identical as polynomials (in the ring Q[x]) so evaluation of x on any field containing Q likewise.
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On September 11 2010 06:04 Assault_1 wrote:whats the domain of x, rational or real?
i would say, think of it as being the rationals.
but yeah, i am finding this quite hard, though I find the result very interesting.
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I've noticed that threads like this pop up during the school year...
ITT: TL does your math homework for you
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On September 11 2010 06:16 sputnik.theory wrote:
I've noticed that threads like this pop up during the school year...
ITT: TL does your math homework for you
Rest assured, I've already solved this. It's not a homework.
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May I (a non-math major) ask for an example of an f(x) and g(x) polynomial for which this is true.
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On September 11 2010 06:04 Assault_1 wrote:whats the domain of x, rational or real?
I am guessing the real or complex is fine. Once we can prove this for the real it automatically implies that it's also true for the complex.
On September 11 2010 06:18 Seth_ wrote:
May I (a non-math major) ask for an example of an f(x) and g(x) polynomial for which this is true.
Sure. f(x) = x^2. g(x) = (2x)^2.
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f(Q) clearly equals g(Q) if f(x)=g(ax+b)
Thus WLOG f and g have integer coefficients that are relatively prime and 0 as their constant terms.
Suppose f(x)=a_nx^n+...a_1*x.
Suppose g(x)=b_nx^n+...+b_1*x.
Let p^r be a prime power dividing a_1 but not b_1, if such a prime power exists. Then f(x) will always be divisible by p either less than or equal to 0 times or more than r times, while g(p) is divisible by p at least once and at most r times.
So this means that f(Q)=g(Q) then a_1= plus/minus b_1
One can continue inductively through all the coefficients
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Or something like that... don't have time to check everything but should be close 
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On September 11 2010 06:25 Muirhead wrote:
f(Q) clearly equals g(Q) if f(x)=g(ax+b)
Thus WLOG f and g have integer coefficients that are relatively prime and 0 as their constant terms.
Suppose f(x)=a_nx^n+...a_1*x.
Suppose g(x)=b_nx^n+...+b_1*x.
Let p^r be a prime power dividing a_1 but not b_1, if such a prime power exists. Then f(x) will always be divisible by p either less than or equal to 0 times or more than r times, while g(p) is divisible by p at least once and at most r times.
So this means that f(Q)=g(Q) then a_1= plus/minus b_1
One can continue inductively through all the coefficients
Need a lot more details, and I'm not convinced of this argument either. You're saying WLOG assume 0 as their constant terms? And then you're assuming deg f = deg g (this is true, but you didn't mention how to prove this at all). And then there's the next paragraph which doesn't seem to make much sense to me.
Enjoy your Friday anyway! 
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Here's an attempt for any field, F:
+ Show Spoiler +
Since f(F)=g(F), g(F) includes f(x) and there must be an element s of F where g(s)=f(x).
Rewrite s in the form ax+b for some a,b in F:
If x = 0, let b = s and a be any number in F.
Then ax+b = a 0 + s = s.
If x is not 0, let a = x^-1, the multiplicative inverse of x in F, and b = s - 1.
Then ax+b = (x^-1) x + (s - 1) = 1 + (s - 1) = s.
In both cases, ax+b = s and g(ax+b) = g(s) = f(x).
Edit: Never mind, I think I misread the problem.
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EPT
