Getting to Cambridge: STEP by STEP!

You did a putnam question? - Sure, no problemo (if I can even wrap my head around it... )

haha well it seemed a little challenging but wasnt too bad, isnt it the Americans version of STEP/ Olympiad sort of stuff?

It's aimed at American undergraduates.

oh haha well I thought it was strange they required you to know about Cayley-Hamilton theorem (atleast I needed it for my solution, perhaps missed an easier way)

EnglishMuon at King's: Zain, mate - the worst thing happened. I was writing my shopping list and accidentally solved the Riemann Hypothesis. fml why does this happen to me

talking about doing maths, please would you look over my solution to a putnam question later?

It's aimed at American undergraduates.

wtf dude

I dont think it was one of the hard ones. It helped that I was talking to Zain about some of the theorems I used only a couple days ago

Talking about doing maths, please would you look over my solution to a Putnam question later?

I will have a look. I like putnam ****


Posted from TSR Mobile

Thanks ill write it up in a min

Wat problem is it.


Posted from TSR Mobile

I dont think it was one of the hard ones. It helped that I was talking to Zain about some of the theorems I used only a couple days ago

Yeah there are some not-that-difficult Putnam problems but most are acknowledged to be very difficult. I've done a few of the easier ones since my Calculus text is American . (I've therefore done quite a few AIME problems and the like, which are really fun .)

ah nice. But yeah that problem still seems hard to get a full refined solution. Ive got the general thing down on paper now but its far from floor proof with 0 assumptions . Which calculus text do you use btw?

Well here is some of my working so far so hopefully it isnt completely wrong!
B3

Let $\mathbf{M} = \begin{pmatrix} a & {a+d} \\ {a+2d} & {a+3d} \end{pmatrix}$
By Cayley-Hamilton theorem, M satisfies its own characteristic polynomial (if thats the correct term for it):
$| \mathbf{M} - \lambda \mathbf{I} | = \lambda ^{2} - (2a+3d) \lambda -2d^{2}$ on simplification. Hence $\mathbf{M} ^{2} - (2a+3d) \mathbf{M} -2d^{2} \mathbf{I} = \mathbf{0}$

Now this is the part that maybe could be written better but idk.

Let $d=0$ so $\mathbf{M} = \begin{pmatrix} a & {a} \\ {a} & {a} \end{pmatrix} , \mathbf{M}^{2} = 2a \mathbf{M}$. Then clearly every power of M also has elements of an arithmetic progression 0, hence $\mathbf{M} = \begin{pmatrix} a & {a} \\ {a} & {a} \end{pmatrix} \in S$

Now let $d= - \frac{2a}{3}$ : $\mathbf{M} = \begin{pmatrix} a & { \frac{a}{3}} \\ {- \frac{a}{3}} & {-a} \end{pmatrix}, \mathbf{M}^{2} = 2d^{2} \mathbf{I}$. Consider M^3: $\mathbf{M}^{3}= \begin{pmatrix} {2ad^{2}} & {2ad^{2} + 2d^{3}} \\ {2ad^{2} + 4d^{3}} & {2ad^{2} + 6d^{3}} \end{pmatrix}$ which also has 'arithmetic elements' so $\begin{pmatrix} a & { \frac{a}{3}} \\ {- \frac{a}{3}} & {-a} \end{pmatrix} \in S$.

I think these two types of matrix satisfy all of the elements of our set S, so Im now trying to prove that all our elements take the form. I havent got a solution yet but Ive got a couple of ideas Im looking in to, but not sure if I know enough about vector spaces to do it properly or not. I was thinking something along the lines of writing our general solution as $\mathbf{M}^{k} = ( \begin{pmatrix} a & {a+d} \\ {a+2d} & {a+3d} \end{pmatrix})^{k}= ( a \begin{pmatrix} 1 & {1} \\ {1} & {1} \end{pmatrix} + d \begin{pmatrix} 0 & {1} \\ {2} & {3} \end{pmatrix})^{k} \in S$ but not convinced of the follow up (if there even is one).

problem B3 on 2015 paper