Official TSR Mathematical Society
Maths and statistics discussion, revision, exam and homework help.
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ok this one is too simple....(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.
3^5 = 243 = 1 (mod 121)
3^10 = 1^5 (mod 121) -
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u can use double intergal, transform to polar, and the answer comes out nicely...(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). -
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Was this solution correct?(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.
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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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.(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?
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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I was told that you have to request to join a society at(Original post by Spenceman_)
So how do we actually get the image next to our usernames?
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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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.(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. -
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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.(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. -
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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?(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.
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?
