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Official TSR Mathematical Society

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    (Original post by J.F.N)
    Alright, maybe that one is too difficult. Lets start with a simpler one: prove that 3 is a Fermat solution with respect to 11.
    ok this one is too simple....

    3^5 = 243 = 1 (mod 121)
    3^10 = 1^5 (mod 121)
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    Ok, here is another interesting question: find the integral of e^(-x^2) from -infinity to +infinity.

    (Hint: you can use Taylor series. There is also a cool way of solving it using two variables).
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    (Original post by J.F.N)
    Ok, here is another interesting question: find the integral of e^(-x^2) from -infinity to +infinity.

    (Hint: you can use Taylor series. There is also a cool way of solving it using two variables).
    u can use double intergal, transform to polar, and the answer comes out nicely...
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    Does the TSR Mathematical Society need an image so we all have one next to our usernames? (eg like the photographers society, the political parties, CHAT etc)
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    Yeh! It sure does! I second that.
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    I think Euler's onto it.
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    see attached
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    (Original post by Galois)
    If you argue that the cosmonaut is moving at around 25000 mph (Orbital velocity - Kepler's 3rd law) in an elliptical orbit with the centre of the earth at one focus of the ellipse, then you can evidently say that the velocity that could be given to the lens cap is so small relatively that it will move in a slightly different elliptical orbit around the earth.

    Galois.
    Was this solution correct?

    How do you know that the lens cap will orbit the earth at all? Doesn't it also depend on its angle of entry?
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    So how do we actually get the image next to our usernames?
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    (Original post by endeavour)
    Was this solution correct?

    How do you know that the lens cap will orbit the earth at all? Doesn't it also depend on its angle of entry?
    You are told that the cosmonought drops the lens cap towards the Earth - that information is sufficient, the angle is negligible. At those speeds, the orbit of the lens cap will hardly differ from the orbit of the cosmonought.

    This question appeared on a Plus sheet. I emailed the answer to the Professor involved and he seemed to be happy with it.
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    (Original post by Spenceman_)
    So how do we actually get the image next to our usernames?
    I was told that you have to request to join a society at

    http://www.thestudentroom.co.uk/pro...=editusergroups

    However the maths society hasn't been approved yet, it's still waiting to be activated by an admin. Then you can get the picture up near your name.
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    A is putting society approval on hold for a bit cos of server issues. Shouldn't be long though.
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    Ah I see.
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    How do i join?
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    (Original post by rpotter)
    How do i join?
    Right at the beginning of the thread he said PM him i.e. Euler if u want to join the group.
    He is offline atm so u will have to wait until he's online to be part of this group.
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    Its been dead around here for a while. Here's a new question: find all sets of three primes such that the sum of any two of them is a perfect square.
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    (Original post by J.F.N)
    Its been dead around here for a while. Here's a new question: find all sets of three primes such that the sum of any two of them is a perfect square.
    At first glance (with some basic modular arithmetic) I don't think there are any - at least sets of three distinct primes - but I was rushed and maybe talkng rubbish. Of course you might wish to include (2,2,2) as a possible answer.
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    (Original post by RichE)
    At first glance (with some basic modular arithmetic) I don't think there are any - at least sets of three distinct primes - but I was rushed and maybe talkng rubbish. Of course you might wish to include (2,2,2) as a possible answer.
    There are solutions which have distinct primes. Anyway, there are quite a few sets of the form {2,2,x} where x=m^2 -2 for some integer m... Of course, this does not always give a prime (e.g. 11^2-2=119=7*17), but it does generate a prime a whole lot of the time.
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    can i join the maths society anyone?
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    (Original post by J.F.N)
    There are solutions which have distinct primes. Anyway, there are quite a few sets of the form {2,2,x} where x=m^2 -2 for some integer m... Of course, this does not always give a prime (e.g. 11^2-2=119=7*17), but it does generate a prime a whole lot of the time.
    Yes of course I can see solutions like {2,2,7} and there may be infinitely many such triples (not sure how you'd show this). But I don't see how there could be three distinct primes. My reasoning went as follows, and just uses simple modular arithmetic. Can you give me a counter-example to it because I don't see its flaw?

    Let p<q<r be three primes such that

    p+q = a^2
    p+r = b^2
    q+r = c^2

    Suppose (for a contradiction) that p,q,r are all odd. Then a^2,b^2,c^2 are all even - so a,b,c are all even. But then a^2,b^2,c^2 are all 0 mod 4.

    If p = 1 mod 4 then q and r are 3 mod 4 and so q+r is 2 mod 4 - contradiction.
    If p = 3 mod 4 then q and r are 1 mod 4 and so q+r is 2 mod 4 - contradiction.

    So they cannot each be odd and hence p = 2.

    Now we're looking to solve

    2+q = a^2
    2+r = b^2
    q+ r = c^2

    As the primes are distinct then q and r are odd and hence so are a^2 and b^2, whilst c^2 is even. Therefore a^2 and b^2 are 1 mod 4 and c^2 is 0 mod 4. (Square numbers are either 0 or 1 mod 4.)

    So q and r are both 3 mod 4. Hence q+r is 2 mod 4 which contradicts the earlier proof that it was 0 mod 4.

    QED - no such distinct triple of primes exists.

    Can you spot the error? :confused:

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Updated: May 28, 2014
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