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    (Original post by Slumpy)
    Can't you do that in about one step? 1^2=3^2=5^2=7^2=1mod8, so the only a s.t. a^k=7mod8 for some k is 7.
    Indeed! :lol: Now I just can't remember why I took k<l,m without loss of generality yesterday. :beard:
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    (Original post by und)
    Indeed! :lol: Now I just can't remember why I took k<l,m without loss of generality yesterday. :beard:
    The order is irrelevant, so by relabeling you can guarantee that k<l,m.
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    (Original post by Slumpy)
    The order is irrelevant, so by relabeling you can guarantee that k<l,m.
    I know but what about k=l<m. I'm sure I had good reasons!
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    Solution 45

    p^k+p^l+p^m=n^2

    Clearly we cannot have k=l=m because this would imply that n had a factor of 9 and hence that 3|p. Consider k&lt;m,l. By writing n^2=p^{k}q^2 and dividing, we get:

    1+p^{l-k}+p^{m-k}=q^2 \Rightarrow p^{l-k}+p^{m-k}=(q+1)(q-1)

    Since p is odd, the right hand side is even and is a multiple of 2^3. Therefore in the case where l \neq m (with equality 4 does not divide the right hand side, so this is impossible) where we can assume without loss of generality that m-l \geq 1, p^{m-l}=-1 \mod 8.

    It can be shown by exhaustion that p^{m-l}=-1\mod 8 for some m-l \geq 1 if, and only if, p=-1\mod 8.

    Now consider k=l&lt;m\Rightarrow p^{m-l}=q^2-2. Clearly q is odd, and by exhaustion it can be shown that the residue of (2r+1)^2-2=-1\mod 8, completing the proof.
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    (Original post by und)
    Solution 45 (almost complete)

    p^k+p^l+p^m=n^2

    Clearly we cannot have k=l=m because this would imply that n had a factor of 9 and hence that 3|p. Furthermore, ... Therefore, without loss of generality we can assume k&lt;m,l. By writing n^2=p^{k}q^2 and dividing, we get:

    1+p^{l-k}+p^{m-k}=q^2 \Rightarrow p^{l-k}+p^{m-k}=(q+1)(q-1)

    Since p is odd, the right hand side is even and is a multiple of 2^3. Therefore in the case where l \neq m (with equality 4 does not divide the right hand side, so this is impossible) where we can assume without loss of generality that m-l \geq 1, p^{m-l}=-1 \mod 8.

    It can be shown by exhaustion that p^{m-l}=-1\mod 8 for some m-l \geq 1 if, and only if, p=-1\mod 8, completing the proof.
    Doesn't k have to be even for this to be true?
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    (Original post by metaltron)
    Doesn't k have to be even for this to be true?
    If k is odd then n is not an integer.
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    (Original post by und)
    ...
    Excellent. The last step can be done by contradiction.
    Spoiler:
    Show
    Assume m-l \equiv 0 \pmod2. Thence 1 \equiv -1 \pmod8 - contradiction. Hence m-l \equiv 1 \pmod2 and p^{m-l} \equiv p \equiv -1 \pmod8.
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    (Original post by und)
    If k is odd then n is not an integer.
    Of course! Well done.
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    This has a really neat little solution. And it is a good demonstration of how useful modular arithmetic can be.

    Problem 49**

    Show that  7| 5555^{2222} + 2222^{5555}
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    Solution 49

    We have 2222^{5555}+5555^{2222} \equiv 3^{5}+4^{2} \equiv 0 \pmod7.
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    (Original post by Mladenov)
    Excellent. The last step can be done by contradiction.
    Spoiler:
    Show
    Assume m-l \equiv 0 \pmod2. Thence 1 \equiv -1 \pmod8 - contradiction. Hence m-l \equiv 1 \pmod2 and p^{m-l} \equiv p \equiv -1 \pmod8.
    I still need to prove it for k=l&lt;m. :beard:
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    (Original post by Mladenov)
    Solution 49

    We have 2222^{5555}+5555^{2222} \equiv 3^{5}+4^{2} \equiv 0 \pmod7.
    What are the conditions for replacing everything with its modular counterpart? I'm not that familiar with modular arithmetic so I should probably read up on it a little.
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    (Original post by und)
    I still need to prove it for k=l&lt;m. :beard:
    Not really. Assume k=l&lt;m. Then p^{m-l}+2 is a perfect square (why?). Hence  p^{m-l} \equiv -1 \pmod 8.

    (Original post by und)
    What are the conditions for replacing everything with its modular counterpart? I'm not that familiar with modular arithmetic so I should probably read up on it a little.
    Those are relatively prime to 7. Use Fermat's little theorem and \pmod 6
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    (Original post by Mladenov)
    Not really. Assume k=l&lt;m. Then p^{m-l}+2 is a perfect square (why?). Hence  p^{m-l} \equiv -1 \pmod 8.
    It's clear that it's a perfect square but the second part isn't that clear in my opinion so I'll add it in.
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    Problem 50:

    An inextensible rope of length l and uniform mass per unit length lies on a rough table with one end on an edge. The co-efficient of friction between the table and the rope is μ. The rope receives an impulse which sets it moving off of the edge of the table at speed v. Prove that, if the rope does not fall off the table, then:

     v^2&lt;\dfrac{lg\mu^2}{1+\mu }

    By considering this as  \mu varies between zero and 1, find the maximum possible impulse that could potentially be given to the rope without it falling off of the table.
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    (Original post by DJMayes)
    Problem 50:

    An inextensible rope of length l and uniform mass per unit length lies on a rough table with one end on an edge. The co-efficient of friction between the table and the rope is μ. The rope receives an impulse which sets it moving off of the edge of the table at speed v. Prove that, if the rope does not fall off the table, then:

     v^2&lt;\dfrac{lg\mu^2}{1+\mu }

    By considering this as  \mu varies between zero and 1, find the maximum possible impulse that could potentially be given to the rope without it falling off of the table.
    Having solved the equation and obtained something not very nice, I don't feel inclined to continue!
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    Problem 51**

    One of my favorite number theory problems.

    Let n &gt;1 be an integer, and k be the number of distinct prime divisors of n. Then there exists an integer a, 1&lt;a&lt;\frac{n}{k}+1, such that a^{2} \equiv a \pmod n.

    Problem 52**

    Find all infinite bounded sequences a_{1},a_{2},... of positive integers such that for each n&gt; 2, a_{n}=\frac{a_{n-1}+a_{n-2}}{(a_{n-1},a_{n-2})}.

    Problem 53**

    For those who have not covered much number theory.

    Let k be a positive integer. Then there exist infinitely many positive integers n such that n2^{k}-7 is a perfect suqare.
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    (Original post by und)
    Having solved the equation and obtained something not very nice, I don't feel inclined to continue!
    :lol:

    If you want to check solutions then you can PM me; mine may well need checking as well.
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    (Original post by ukdragon37)
    A little, yes. But then topos theory is quite a different beast to set theory.
    In the Cambridge 4th year some people (tried) to run a Topos Theory reading group. I think in the end only about 5 or so people finished the course.

    I started it and then gave up after about 6-7 'lectures' because it was just utterly incomprehensible.
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    (Original post by DJMayes)
    :lol:

    If you want to check solutions then you can PM me; mine may well need checking as well.
    This didn't exactly inspire confidence:

    \displaystyle x=\left( \frac{\mu l}{2g(\mu +1)}+\frac{v}{2\sqrt{\frac{g}{l}  (\mu +1)}} \right) e^{\sqrt{\frac{g}{l}(\mu +1)}t} + \left( \frac{\mu l}{2g(\mu +1)}-\frac{v}{2\sqrt{\frac{g}{l}(\mu +1)}} \right) e^{-\sqrt{\frac{g}{l}(\mu +1)}t}-\frac{\mu l}{g(\mu +1)}
 
 
 
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