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$f(x)$ is a polynomial function with integer coefficients satisfying $f(1)=5$ and $f(2)=7$. What is the smallest possible positive value of $f(12)$?

I have no idea on how to begin with this question. Also, I can't get any clue on the significance of integer coefficients.

Soham
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  • Absent such a condition, there is no minimum since Lagrange Interpolation will give rational coefficients if $f(12)\in \mathbb Q$. For instance, if you want $f(12)=-10^6$, we could have $\frac {-999862}{55} + \frac {3000301}{110 } x- \frac {1000027}{110}x^2$. – lulu Oct 11 '18 at 18:28
  • @lulu Edit: smallest possible positive value of $f(12)$ – Soham Oct 11 '18 at 18:28
  • @lulu, integer coefficents is the keyword and it should be smallest positive value of f(12) – Love Invariants Oct 11 '18 at 18:29
  • Ok, well you could still get $f(12)=1$ as $\frac {139}{55} + \frac {149}{55}x - \frac {13}{55}x^2$. – lulu Oct 11 '18 at 18:30
  • @lulu What about integer coefficients? – Soham Oct 11 '18 at 18:31
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    Hint: If $f(x)\in\Bbb{Z}[x]$ then $f(m)-f(n)$ is an integer multiple of $m-n$. – Jyrki Lahtonen Oct 11 '18 at 18:32
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    In your post, you asked what the significance of integer coefficients was. My comments are intended to show that, absent such a condition, there is no restriction on $f(12)$ since Lagrange preserves rationality but not integrality. – lulu Oct 11 '18 at 18:32
  • @lulu Okay... thanks ;-) – Soham Oct 11 '18 at 18:32
  • @JyrkiLahtonen Can you please extend a bit, sir? – Soham Oct 11 '18 at 18:34
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    So $f(12)-f(2)$ must be a multiple of ten and $f(12)-f(1)$ is a multiple of eleven. I guess you are expected to set up the congruence system and solve it. – Jyrki Lahtonen Oct 11 '18 at 18:35
  • @JyrkiLahtonen- How does that lead to the answer? – Love Invariants Oct 11 '18 at 18:36
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    @JyrkiLahtonen $f(12)=10k+7$ and $f(12)=11p+5$. The smallest integers $k,p$ for which this is true gives $f(12)=27$... for $k=p=2$. So, $27$ is the answer. Am I correct? – Soham Oct 11 '18 at 18:39
  • I am fairly sure that we have handled similar questions in the past. Did you search? Anyway, solving that system of congruences, Chinese remainder theorem style, will give you a set of possible values of $f(12)$. Pick the smallest positive one. Then construct a polynomial $f(x)$ achieving that. – Jyrki Lahtonen Oct 11 '18 at 18:39
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    Yes, Tatan. That is correct. Still, that congruence is only a necessary condition. Can you find a polynomial $f(x)$ such that $f(1), f(2)$ and $f(12)$ all have the prescribed value? – Jyrki Lahtonen Oct 11 '18 at 18:40
  • And... Do you see where that divisibility criterion comes from? – Jyrki Lahtonen Oct 11 '18 at 18:41
  • @JyrkiLahtonen I am not too confident with congruences. I don't know much of Chinese Remainder theorem – Soham Oct 11 '18 at 18:43
  • @tatan- But how could you say that f(12) will assume 27 for some value. You have to prove. For eg if it were given that f(5)=14 then what would you do. – Love Invariants Oct 11 '18 at 18:44
  • @JyrkiLahtonen- 2x+3. – Love Invariants Oct 11 '18 at 18:44
  • Correct, @LoveInvariants :-) – Jyrki Lahtonen Oct 11 '18 at 18:45
  • @LoveInvariants I used the facts that Jyrki sir said... that divisibility condition – Soham Oct 11 '18 at 18:45
  • @LoveInvariants How did you get that? You solved the system of congruences? – Soham Oct 11 '18 at 18:46
  • @tatan- yup, for this question, you can quote the polynomial 2x+3 easily with the solution but if it is too complicated. How would you prove that f(12) will assume certain value. – Love Invariants Oct 11 '18 at 18:48
  • @JyrkiLahtonen Can you provide a link or add an answer to explain in a bit more detai? – Soham Oct 11 '18 at 18:49
  • @tatan- I was solving for positive integral coefficients though :P. Subtracted both functions to get 2 as the coefficient of x. – Love Invariants Oct 11 '18 at 18:49
  • Looking for one, but couldn't find a good one right away. I summoned Bill Dubuque, for IIRC he has explained it somewhere here. – Jyrki Lahtonen Oct 11 '18 at 18:50
  • @JyrkiLahtonen Thanks, looking forward to the solution ;-) – Soham Oct 11 '18 at 18:52
  • Does this help? – Jyrki Lahtonen Oct 11 '18 at 18:53
  • @JyrkiLahtonen Thanks. I need some time. I would definitely get back on this in some time ;-) – Soham Oct 11 '18 at 18:56
  • By any chance do you have the final answer to the question? I think the mininum to $f(12)$ is $5$. – Zacky Oct 11 '18 at 19:36
  • @Dahaka Answer says its $27$ but I am ready to accept $5$ if its logical – Soham Oct 11 '18 at 19:58
  • Oh, I made a few mistakes anyway. Well, then here is a thought: Since $f(1)=5=2(1)+3$ and $f(2)=7=2(2)+3$ we have that $f(x) - (2x+3)$ has as roots $1 $ and $2 $, also the function is of the form: $$f(x) =g(x) (x-1)(x-2)+2x+3\rightarrow f(12)=g(12)\cdot 11\cdot 10+27$$ Where $g(x) $ is a random polynomial with integer coefficients and with atleast $2$ degree less than $f(x) $. Now we only need to show that the minimum of this polynomial $g(x) $ in $x=12$ is $0$ and we obtain that the minimum of $f(12)$ is $27$. I don't really know for the moment how to show that, hopefully someone can help. – Zacky Oct 11 '18 at 20:29
  • @Dahaka: You are practically done; $g(12)$ is an integer, and so if it is negative you have an upper bound on it. No need to care about degree. The core structure of this proof is the same as Jryki's. – user21820 Oct 12 '18 at 07:37

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