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    Alternative (1) Solution 526:

    We know that \displaystyle (\sqrt{2})^{\log_{\sqrt{2}}3 } = 3 and we know that \sqrt{2} is irrational. This means taht we need only prove that \log_{\sqrt{2}} 3 is irrational.

    We can do so easily, since \displaystyle \log_{\sqrt{2} 3} = \frac{\log_2 3}{\\log_2 \sqrt{2}} = 2 \log_2 3. Now assume that this equals \frac{m}{n} where \gcd{m,n} = 1 then we have \displaystyle \log_2 3 = \frac{m}{2n} so that 3^{2n} = 2^m, contradiction so \log_{\sqrt{2}} 3 is irrational and we are done.
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    A little basic number theory pilfered from Putnam.

    Problem 537: **
    Show that for each p>17, where p is prime.
    16320|p^{32}-1.

    Problem 538: **
    Show that there are no integer solutions x,y to:
    (x+1)^n-x^n=ny where n \in \mathbb{N} \setminus \{1 \}.
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    (Original post by joostan)
    A little basic number theory pilfered from Putnam.

    Problem 537: **
    Show that for each p>17:
    16320|p^{32}-1.

    Problem 538: **
    Show that there are no integer solutions x,y to:
    (x+1)^n-x^n=ny where n \in \mathbb{N} \setminus \{1 \}.
    What's p? A prime?
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    (Original post by Renzhi10122)
    What's p? A prime?
    Yarp, p's a prime.
    Probably shoulda specified, but I'd say it's reasonably clear from context.
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    (Original post by joostan)
    Yarp, p's a prime.
    Probably shoulda specified, but I'd say it's reasonably clear from context.
    Fair enough

    Solution 537

    Firstly note that  16320=2^6*3*5*17 . By the LTE lemma, we have
     v_2(p^{32}-1)=v_2(p-1)+v_2(p+1)+v_2(32)-1\geq 3+5-1=7 , where  v_2(n) is the greatest power of 2 that divides  n . Therefore,  p^{32}-1 is divisible by  2^7 , a stronger result. Now note that  p is not divisible by 3,5, or 17, since it is prime. By Fermat's Little Theorem,  a^{k(q-1)}\equiv 1  (mod  q) . Noting that each of 3, 5 and 17 can be written in the form  2^m +1 , in particular, for  m\leq 5 , setting  q as each of the three primes, we see that 32 is a multiple of  q-1 , and so  p^{32}-1 is divisible by 3, 5 and 17.
    Thus,  16320|p^{32}-1 and indeed  32640|p^{32}-1 for all primes  p\neq 2,3,5,17 .

    EDIT: We could've used difference of two squares for the first bit, but only thought of that afterwards
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    Solution 537 (2)
    Damn I really was a bit slow :-)
    Spoiler:
    Show

    First note that  16320 = 2^6 \times 5 \times 3 \times 17
    If p is a prime greater than 17 then p is odd and not divisible by 3, hence  p \equiv 1 (mod 3) therefore  p^{32} \equiv 1 (mod 3) and hence  p^{32} - 1 \equiv 0 (mod 3) , or  p \equiv 2 (mod 3) , in which case since p^2 \equiv 4 \equiv 1 (mod 3) and hence  p^{32} \equiv 0 So  3|p^{32} - 1 .

    Note that we can rewrite the expression as  (p-1)(p+1)(p^2 +1)(p^4 + 1)(p^8 + 1)(p^{16} + 1) . Each factor is even since p is odd and hence any power of p is odd. Hence  64 |p^{32} - 1

    Finally let us consider the divisor 17. If p is a prime greater than 17 then p and 17 are coprime. Hence by Fermat's Little Theorem

    p^{16} \equiv 1 (mod 17) and hence p^{32} \equiv 1 (mod 17) so p^{32} - 1 \equiv 0 (mod 17)

    So 3, 5, 64 and 17 divide p^{32} - 1 . Hence 16320 divides p^{32} - 1.





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    (Original post by Renzhi10122)
    x
    (Original post by 16Characters....)
    x
    Impressive, both of you. My number theory is ****e... :rofl:

    Protip: To typeset a \equiv b \pmod c use a \equiv b \pmod c instead of a \equiv b (mod c) since that just squishes the variables m, o, d together.

    Edit: It evens saves you the hassle of putting the brackets around the mod and it just requires a slash and a p before the mod.
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    Political Ambassador
    Some more analysis

    Problem 539

    Let f:[0,1] \rightarrow [0,1] be any function that is continuous on [0,1]

    Prove that there exists a constant k \in [0,1] such that

    f(k) = \sin\left(\dfrac{k\pi}{2}\right)


    Posted from TSR Mobile
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    (Original post by Indeterminate)
    Some more analysis

    Problem 539

    Let f:[0,1] \rightarrow [0,1] be any function that is continuous on [0,1]

    Prove that there exists a constant k \in [0,1] such that

    f(k) = \sin\left(\dfrac{k\pi}{2}\right)


    Posted from TSR Mobile
    Solution 539.

    Suppose that f(x)&gt;sin(\frac{x\pi}{2}) for all x\in [0,1]. Then we must have f(1)&gt;sin(\frac{\pi}{2})=1, which is impossible since this states f(1) is not in the codomain. A similar argument at x=0 shows we cannot have f(x)<sin(x*pi/2) for all x in [0,1] either.

    Thus there must exist a\in [0,1] for which f(x)\geq \sin(\frac{x\pi}{2}), and b\in [0,1] such that f(x)\leq \sin(\frac{x\pi}{2}). Suppose without loss of generality we have a\leq b. Let  g(x)=f(x)-sin(\frac{x\pi}{2}). Then at a, g(a)\geq 0 and at b g(b)\leq 0. Thus by the intermediate value theorem, there exists a c such that  a\leq c \leq b for which g(c)=0 , which implies by construction of g  f(c)=\sin(\frac{c\pi}{2})

    Edit: A nice intuitive picture for how this theorem works: Draw a square, and from the bottom left corner to the top right corner, draw a curve resembling the graph of sin from 0 to pi/2. Now try and join the left side to the right side with a continuous curve that doesn't intersect the sine curve .
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    (Original post by FireGarden)
    Solution 539.

    Suppose that f(x)&gt;sin(\frac{x\pi}{2}) for all x\in [0,1]. Then we must have f(1)&gt;sin(\frac{\pi}{2})=1, which is impossible since this states f(1) is not in the codomain. A similar argument at x=0 shows we cannot have f(x)<sin(x*pi/2) for all x in [0,1] either.

    Thus there must exist a\in [0,1] for which f(x)\geq \sin(\frac{x\pi}{2}), and b\in [0,1] such that f(x)\leq \sin(\frac{x\pi}{2}). Suppose without loss of generality we have a\leq b. Let  g(x)=f(x)-sin(\frac{x\pi}{2}). Then at a, g(a)\geq 0 and at b g(b)\leq 0. Thus by the intermediate value theorem, there exists a c such that  a\leq c \leq b for which g(c)=0 , which implies by construction of g  f(c)=\sin(\frac{c\pi}{2})
    All fine, but a bit verbose:

    Define g(x) = sin(pi x / 2) - f(x). Then g(0) = -f(0), g(1) = 1- f(1) and so since we know 0<=f(x)<=1, g(0)<=0, g(1)>=0. If we have equality in either case we're done, otherwise by IVT we can find k in (0,1) with g(k) = 0 and we're done.

    [Questions like this come up often enough that you should make sure you know how to answer them concisely. The defn of IVT I use requires f(a) < 0, f(b) > 0 (i.e. it doesn't allow equality) or it could go a little shorter still].
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    Problem 540 (**/***)

    a) Show that:

    \displaystyle \int_0^{2\pi} \frac{r^2\sin^2 2\theta}{1-2r^2\cos 2\theta+r^4} \ d\theta = \pi r^2

    stating any conditions that must hold for this to be valid.

    b) Find the value of \displaystyle \int_0^{2\pi} \frac{4\sin^2 2\theta}{17-8\cos 2\theta} \ d\theta

    (Hints available if necessary)
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    (Original post by atsruser)
    Problem 531: */**

    Give a geometrical argument to show that, if x^2+y^2=1, then

    \displaystyle \int_{\frac{\sqrt{2}}{2}}^{ \frac{\sqrt{2+\sqrt{2}}}{2} } \frac{1}{\sqrt{1-x^2}} \ dx = \frac{\pi}{8}
    Bumping with hints:

    Spoiler:
    Show
    Hint 1)

    Note that since x^2+y^2=1, we have \int \frac{1}{\sqrt{1-x^2}} \ dx = \int \frac{dx}{y}.

    Which property of the curve does \frac{dx}{y} measure?

    Spoiler:
    Show

    dx is a fixed, small increment in x; how does \frac{dx}{y} vary as x varies from 0 to 1?; which property of the curve varies with dx in a similar way?

    Hint 2)

    To which points on the curve do the limits correspond? What is the significance of the values of the limits, and what is their relationship to the quantity measured by \frac{dx}{y}?

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    (Original post by atsruser)
    x
    That's quite an interesting problem, unfortunately, it's three in the morning here and I'm heading off to bed, but I'll give these a try first thing in the morning, thanks for your contribution!

    I can't quite wrap my head around the whole \frac{\mathrm{d}x}{y} thing.
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    Can't believe I've never used this thread before.


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    UProblem 541
    I am on a IPad so not really sure about this bold stuff etc.
    Prove the for natural n the fraction (21n+4)/(14n+3) is irreducible.
    I will post a solution later on today.

    This is a problem from an old IMO problem!
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    Solution 541:
    Observe that 3(14n+3)=2(21n+4)+1.
    Thus if p|(14n+3) and p|(21n+4) for p \in \mathbb{N} then p|(3(14n+3)-2(21n+4)) so p|1 \Leftrightarrow p=1.
    Hence we have 14n+3 and 21n+4 coprime, so:
    \dfrac{21n+4}{14n+3} is irreducible.
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    (Original post by joostan)
    Solution 541:
    Observe that 3(14n+3)=2(21n+4)+1.
    Thus if p|(14n+3) and p|(21n+4) for p \in \mathbb{N} then p|(3(14n+3)-2(21n+4)) so p|1 \Leftrightarrow p=1.
    Hence we have 14n+3 and 21n+4 coprime, so:
    \dfrac{21n+4}{14n+3} is irreducible.
    Mine essentially the same but slight different at the start I think.
    Gcd(21n+4,14n+3)=gcd(7n+1,14n+3)
    =gcd(7n+1,14n+3-2(7n+1))
    =gcd(7n+1,1)=1 Q.E.D


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    (Original post by physicsmaths)
    Mine essentially the same but slight different at the start I think.
    Gcd(21n+4,14n+3)=gcd(7n+1,14n+3)
    =gcd(7n+1,14n+3-2(7n+1))
    =gcd(7n+1,1)=1 Q.E.D


    Posted from TSR Mobile
    Sorry to be a buzzkill, but it's specified in the OP to Latex all solutions/problems (I understand that you're on an iPad, but it's not hard to use Latex) - just to maintain austerity.

    Also - it's fine to post a solution to your own problem once somebody else has already answered it, but not before.
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    (Original post by 16Characters....)
    Problem 508

     M is an N x N matrix with N distinct eigenvalues all in the range  - 1 &lt; \lambda &lt; 1 . Prove that:

     I + \displaystyle \sum_{r = 1}^{\infty} M^r

    Converges. {NB: I denotes the N x N identity matrix}
    Bumping with a hint:

    Spoiler:
    Show

    A matrix with all of its eigenvalues distinct is diagonalisable.
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    (Original post by 16Characters....)
    Problem 508

     M is an N x N matrix with N distinct eigenvalues all in the range  - 1 &lt; \lambda &lt; 1 . Prove that:

     I + \displaystyle \sum_{r = 1}^{\infty} M^r

    Converges. {NB: I denotes the N x N identity matrix}
    Let S=I+\displaystyle\sum_{r=1}^ { \infty} M^r
    Since M has N distinct eigenvalues, we can write M=P^{-1}DP for some fixed, invertible N \times N matrix P, and where D=diag(\lambda_1, . . .,\lambda_N).
    Notice that:
    M^r=P^{-1}D^rP

\Rightarrow S=I+\displaystyle\sum_{r=1}^ { \infty} P^{-1}D^rP

\Rightarrow S=\displaystyle\sum_{r=0}^ { \infty} P^{-1}diag(\lambda_1^r, . . .,\lambda_N^r)P.
    But |\lambda_i|&lt;1 \Rightarrow \displaystyle\sum_{r=0}^{\infty} \lambda_i^r = \dfrac{1}{1-\lambda_i}.
    (Assuming \lambda_i \not= 0, if so then \displaystyle\sum_{r=0}^{\infty} \lambda_i^r=0 so still converges.)
    Hence we have that:
    S=P^{-1} diag \left(\dfrac{1}{1-\lambda_1}, . . . ,\dfrac{1}{1-\lambda_N} \right)P.
    Thus the sum converges.
 
 
 
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