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Being divisible by p does not imply that you are a composite number. Why can't P(p)=p itself?

1. Does there exist a non-constant polynomial P(x) with integer coefficients such that P(k) is prime for every positive integer k? Justify your response.

Lols, mega procrastination on my behalf. Am supposed to be brushing up on something for a supervisor meeting. These questions are fun though.

You should probably specify the domain of the functional equation you posted Paradoxica. It it the reals? Assuming this is the case (and in fact this working is also valid if the domain is the integers, rationals or complex numbers), then: 6/ Let x=0 and let y=-f(0)+z, so...

7. \\ \\(2n)!^2=4n^2 \prod_{k=1}^{2k-1} k(2n-k)\\ \\ \leq 4n^2 \prod_{k=1}^{2n-1} n^2=4n^{4n}.\\ \\ $where the inequality came from applying AM-GM to each factor in the product.\\ \\ Taking square roots of both sides completes the proof.$

$5. It is clear that $x=-y$ is a solution for any integer $y$. We now assume $x+y\neq 0$ and so we can divide by this to reduce our equation to $x^2+y^2-xy=x+y$. \\ \\ The reason this is promising is that for $x$ and $y$ large, the LHS should grow a lot faster than the RHS and equality should be...

Just went with the immediate idea that popped into my head, didn't take long to tex/compute. If you want a weaker constant than 1/sqrt(3) (like 1/sqrt(2) for instance) then the same method as above gets you immediate results without needing to go several terms into the product before doing the...

LHS%5E2%3D%5Cprod_%7Bk%3D1%7D%5En%20%5Cleft%28%5Cfrac%7B2k-1%7D%7B2k%7D%5Cright%29%5E2%20%5C%5C%20%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%5Cfrac%7B1%7D%7B2%7D%5Ccdot%5Cfrac%7B3%7D%7B4%7D%5Ccdot%5Cfrac%7B3%7D%7B4%7D%5Ccdot%5Cfrac%7B5%7D%7B6%7D%5Ccdot%5Cfrac%7B5%7D%7B6%7D%5Ccdot%5Cfrac%7B7%7D%7B8%...

$1.\\ \\ $LHS+n=\sum_{k=0}^{n-1} \left(1+\frac{1}{n+k}\right)\\ \\ =\sum_{k=0}^{n-1} \frac{n+k+1}{n+k}\\ \\ \leq n\left(\prod_{k=0}^{n-1} \frac{n+k+1}{n+k}\right)^{1/n} = n\cdot 2^{1/n}.$ $

Exactly, although worded slightly funny. The point is that the j-th term on the LHS MUST be divisible by p_j as every other term trivially is. This means that each a_j must be a non-negative integer multiple of p_j. (Because we do not get a prime factor of p_j in this term from multiplying out...

I still think it is. You will probably slap your head when you spot it :).

Good, you have done the harder part of the question :). It remains to determine the exact solutions of g(x)=x but you are super close.

Re: MX2 2016 Integration Marathon A followup from Integrand's question along the same theme: $Find an expression for $\frac{1}{1+x^2}-\sum_{k=0}^n (-1)^k x^{2k}$. $ $Hence show that for $0\leq x \leq 1$ we have $\arctan(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ .$ $Use the above to...

Re: MX2 2016 Integration Marathon Oh lol that was actually my HSC year. I remember seeing it earlier than that even, I swear something like it is in the Cambridge book but I don't have that on me and it's not really important.

Re: MX2 2016 Integration Marathon Bingo :). A classic example of how good IBP is for bounding things that oscillate fast. P.s. with your discussion of g'(0), you have actually only found the limit of g'(x) as x->0, you have not established differentiability of g at the origin, nor have you...

Re: MX2 2016 Integration Marathon Of course. I try to emphasise creativity with syllabus knowledge over recollection of previously known facts from outside syllabus when I write problems. I think this is a far more important thing for a high schooler to develop (it also makes questions fairer...

Re: MX2 2016 Integration Marathon As another remark: I believe I have seen this same method used to prove these bounds for the trig functions in some MX2 material before. In Cambridge or a past paper/trial perhaps? Note you can also phrase the argument in terms of differentiation...

Re: MX2 2016 Integration Marathon Apologies for vagueness, didn't want to hand-hold too much as I think it is a nice thing for a student to discover by him/herself. I will elaborate now :). Yep, Integrand is exactly right. I was referring to the upper and lower bounds coming from the...

Re: MX2 2016 Integration Marathon Very possible with HSC knowledge, especially with the number of given hints.

Re: MX2 2016 Integration Marathon You definitely would have, but anyway you don't need to have heard of or seen them before to do this question.