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Reply 3880
Prove that axm+bync (mod p)ax^m+by^n \equiv c \ (\textrm{mod }p) has the same number of solutions as axm+bync (mod p)ax^{m'}+by^{n'} \equiv c \ (\textrm{mod }p), where m=gcd(m,p1)m' = \textrm{gcd}(m,p-1) and n=gcd(n,p1)n' = \textrm{gcd}(n,p-1).
(edited 11 years ago)
GHOSH is back alright
Let an odd prime number pp be called special when for all integers aa

(ap)=1    a\displaystyle \left( \frac{a}{p}\right) = -1 \iff a is a generator for (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^{\times}.

Prove that pp is special if, and only if, it is a Fermat prime.
Original post by jack.hadamard
Let an odd prime number pp be called special when for all integers aa

(ap)=1    a\displaystyle \left( \frac{a}{p}\right) = -1 \iff a is a generator for (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^{\times}.

Prove that pp is special if, and only if, it is a Fermat prime.



Lat p be a special prime. We know that (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^{\times} is a cyclic group of order p1p-1. Therefore, if aa is a generator of (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^{\times}, then ai≢aj(modp)a^{i} \not\equiv a^{j}(\mod p) for any i,j{1,2,...,p2}i,j \in \{1,2,...,p-2\}, ij i\not=j. Consequently, ordp(a)=p1ord_{p}(a)=p-1 \Rightarrow aa is a primitive root modulo pp. The converse is also true. Now, every primitive root modulo pp is a quadratic non-residue modulo pp. Hence p12=φ(p1)\frac{p-1}{2}=\varphi(p-1). Writing p1p-1 in the form 1mpiai\prod_{1}^{m}p_{i}^{a_{i}} we see that the only possible solution to our equation is p=2n+1p=2^{n}+1 for some positive integer nn, therefore pp is a Fermat prime.
On the other hand, if p is Fermat prime, then it is special, which is rather routine to verify.
Original post by Mladenov
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Yup. You have all the required ingredients. Welcome! :smile:
Reply 3885
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Original post by Mladenov
Lat p be a special prime. We know that (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^{\times} is a cyclic group of order p1p-1. Therefore, if aa is a generator of (Z/pZ)×(\mathbb{Z}/p\mathbb{Z})^{\times}, then ai≢aj(modp)a^{i} \not\equiv a^{j}(\mod p) for any i,j{1,2,...,p2}i,j \in \{1,2,...,p-2\}, ij i\not=j. Consequently, ordp(a)=p1ord_{p}(a)=p-1 \Rightarrow aa is a primitive root modulo pp. The converse is also true. Now, every primitive root modulo pp is a quadratic non-residue modulo pp. Hence p12=φ(p1)\frac{p-1}{2}=\varphi(p-1). Writing p1p-1 in the form 1mpiai\prod_{1}^{m}p_{i}^{a_{i}} we see that the only possible solution to our equation is p=2n+1p=2^{n}+1 for some positive integer nn, therefore pp is a Fermat prime.
On the other hand, if p is Fermat prime, then it is special, which is rather routine to verify.


I thought I was going well by doing C1 and C2 this morning, but now I realise I'm barely past the second rung of the mathematical ladder!
Original post by nohomo
Find all natural numbers nn for which the number of all positive divisors of the number lcm(1,2,...,n)\textrm{lcm}(1,2,...,n) is equal to 2k2^k for some non-negative integer kk.



Let us fix n1n\ge1. Write lcm(1,2,...,n)=1mpiai \textrm{lcm}(1,2,...,n)=\prod_{1}^{m}p_{i}^{a_{i}}. Then, the number of all positive divisors of lcm(1,2,...,n) \textrm{lcm}(1,2,...,n) is d(lcm(1,2,...,n))=1m(ai+1)d(\textrm{lcm}(1,2,...,n))=\prod_{1}^{m}(a_{i}+1), which is power of two when, and only when, for each i{1,2,...,m}i\in \{1,2,...,m\}, ai=2si1a_{i}=2^{s_{i}}-1, where siZ+s_{i}\in \mathbb{Z^{+}}.
Now suppose n9n\ge 9 is a solution to our problem. Then ν2(lcm(1,2,...,n))=2r1\nu_{2}(\textrm{lcm}(1,2,...,n))=2^{r}-1 for some positive integer r2r\ge 2. Consider prime pp such that 22r12<p22r112^{2^{r-1}-2}< p \le 2^{2^{r-1}-1}, which obviously exists due to Bertrand's postulate. Further p222r2<np^{2} \le 2^{2^{r}-2} < n and p3>23.2r16p^{3} > 2^{3.2^{r-1}-6}. For r4r\ge 4, we have 3.2r16>2r3.2^{r-1}-6 > 2^{r} Hence, p2<n<p3p^{2}<n<p^{3}. Therefore, for r4r \ge 4, there are no solutions. Suppose r=3r=3. If nn is a solution, then 128n<256128\le n < 256. There are no solutions in this interval, for, if there were, then ν7(lcm(1,2,...,n))=2\nu_{7}(\textrm{lcm}(1,2,...,n))=2, which is a contradiction. Next, for r=2r=2, we have 9n<169 \le n < 16. No such nn satisfies our problem since, if there were such nn, we would have ν3(lcm(1,2,...,n))=2\nu_{3}(\textrm{lcm}(1,2,...,n))=2 - contradiction.
Now for r=2r=2, n=8n=8 is a solution. The case r=1r=1 gives us the solutions n=1,2,3n=1,2,3.
Therefore, all solutions to our problem are n=1,2,3,8n=1,2,3,8.
(edited 11 years ago)
Original post by nohomo
Let aa and bb be positive integers. Show that if 4ab14ab-1 divides (4a21)2(4a^2-1)^2 then a=ba=b.



I shall prove the following problem from IMO SL 2007, which was also given at our IMO training camp.

Let aa be a positive integer. Then the number (4a21)2(4a^{2}-1)^{2} has a positive divisor of the form 8ab18ab-1 is and only if aa is even.

Suppose there exist aa and bb such that (4a21)24ab1\frac{(4a^{2}-1)^{2}}{4ab-1} is an integer. Now, if (a,b)(a,b) is a solution, so is (b,a)(b,a). Suppose, without loss of generality, (a,b)(a,b) is a pair for which a<ba< b and a+ba+b is minimal. Since, gcd(4ab1,4a)=1\gcd(4ab-1,4a)=1, we can write (4a21)24ab11(mod4a)\frac{(4a^{2}-1)^{2}}{4ab-1}\equiv -1\pmod{4a}. Therefore, there exists kk such than (4a21)2=(4ab1)(4ak1)(4a^{2}-1)^{2}=(4ab-1)(4ak-1). From this equation and the condition a<ba< b, we see that k<ak < a. Therefore, (a,k)(a,k) is a solution for which a+k<a+ba+k< a+b - contradiction.

Writing (4a21)24a(2b)1\frac{(4a^{2}-1)^2}{4a(2b)-1}, we can conclude, from the above problem, a=2ba=2b, which proves our general problem.
Reply 3889
anyone have the AQA S4 statistics mark scheme for June 2012?
Im new to this btw...

Much appreciated,

Thanks! :smile:
Original post by I am Ace
You caught me :rolleyes:
I can think of no other way of computing
x21lnx\displaystyle\int \frac{x^2-1}{lnx}
Which could possibly be a future STEP question, I'd imagine - if there was another method that I can't see




0 1 If it's from to then that's indeed a classic magic differentiation exercise! But here's an alternative:



Original post by L'art pour l'art

0 1 If it's from to then that's indeed a classic magic differentiation exercise! But here's an alternative:





I can't magic differentiate :frown:
Original post by I am Ace
I can't magic differentiate :frown:
I'm assuming you have done this one though? The trickiest thing about magic differentiation is finding what to magic differentiate. :teehee:
Original post by L'art pour l'art
I'm assuming you have done this one though? The trickiest thing about magic differentiation is finding what to magic differentiate. :teehee:


Nope, maybe I should youtube it
Original post by I am Ace
Nope, maybe I should youtube it

Spoiler

Original post by L'art pour l'art

Spoiler



Do I use partial differentiation treating logxlogx as a constant?
Original post by I am Ace
Do I use partial differentiation treating logxlogx as a constant?
Yes.
Original post by L'art pour l'art
Yes.


OK
then,
I(λ)=logx(xλ1)I'(\lambda)= logx (x^{\lambda}-1)
I done this in my head, I hope it's right...
By logarithmic differentiation
Original post by I am Ace
...
I've completed the solution below in parts. If you have never done differentiation under the integral sign before, then try to study this solution perhaps; otherwise, use the spoilers to finish it yourself (asking any questions you might have in either case). But first let's get the differentiation out of the way.

Unparseable latex formula:

\displaystyle \frac{d}{d\lambda}\left( \frac{x^{\lambda}-1}{\log{x}} \right) = \frac{d}{d\lambda}\left( \frac{x^{\lambda}}{\log{x}} \right) = \frac{1}{\log{x}}\frac{d}{d \lambda}\left x^{ \lambda}

-- if we let y=xλy = x^{\lambda} then

logy=λlogx    1ydydλ=logx    dydλ=ylog(x)=xλlogx.\displaystyle \log{y} = \lambda \log{x} \implies \frac{1}{y} \frac{dy}{d\lambda} = \log{x} \implies \frac{dy}{d\lambda} = y \log(x) = x^{\lambda}\log{x}.

That's we have
Unparseable latex formula:

\displaystyle \frac{d}{d \lambda}\left x^{ \lambda} = x^{\lambda}\log{x}

so ddλ(xλ1logx)=1logxxλlogx=xλ.\displaystyle \frac{d}{d\lambda}\left( \frac{x^{\lambda}-1}{\log{x}} \right) =\frac{1}{\log{x}}\cdot x^{\lambda}\log{x} =x^{\lambda}.

Spoiler



If anyone is interested, the double integral way also gets the same general value

Spoiler



(edited 11 years ago)

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