Official TSR Mathematical Society

Scroll to see replies

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)

Reply 161

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).

Reply 162

Euler

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...

Reply 163

k@tie

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)

Reply 164

Spenceman_

Yeh! It sure does! I second that.

Reply 165

Muse

I think Euler's onto it.

see attached

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?

Reply 168

Spenceman_

So how do we actually get the image next to our usernames?

Reply 169

Gauss

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.

Reply 170

Gauss

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.

Reply 171

Muse

A is putting society approval on hold for a bit cos of server issues. Shouldn't be long though.

Ah I see.

How do i join?

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.

Reply 175

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.

Reply 176

RichE

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.

Reply 177

J.F.N

RichE

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.

Reply 178

Font

can i join the maths society anyone?

Reply 179

RichE

J.F.N

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: