TT's Maths Thread

[TrueTears]

Lolol. Differs from person to person I guess. 可谓仁者见仁，智者见智! 

hahaha I see you have been reading my Chinese essays!

[brightsky]

Hehe, sure have! Got to say, they are wayy too pro!!  :p

[EvangelionZeta]

Nope haven't been to him, but from what I've heard from friends, he's pretty shit.

Really?  I've heard he teaches concepts very well (haven't been to his class personally, although I sat his Methods practice exams.  :p).  Also his questions tend to be challenging enough to make actual VCAA exams appear very easy. 

[TrueTears]

Nope haven't been to him, but from what I've heard from friends, he's pretty shit.

Really?  I've heard he teaches concepts very well (haven't been to his class personally, although I sat his Methods practice exams.  :p).  Also his questions tend to be challenging enough to make actual VCAA exams appear very easy.  

ye


Alright just wondering, if we have integers and such that then there exists some integers and such that

Then does that mean if we have integers such that then does there exists some integers such that ?

If there is, then how do we solve the linear diophantine equation for ? Ie, what would the general solution be? (Is this question beyond my abilities? Is there even a general solution?)

Thanks

[zzdfa]

yea it's pretty simple,

try proving it for n=3.
 let g=(a1,a2)

use the fact that
 (a1,a2,a3)= (g,a3)

and


g=a1y1+a2y2 for some y1,y2


the general case should follow easily, by induction.

edit: actually, if you're using induction, it's easier to prove the general case instead:

a1x1+...+anxn=m has a solution iff  (a1,...,an)|m

the inductive proof also shows you how to find the solution.

[TrueTears]

Finally I can do some modular arithmetic questions.

I must say this probs has been the most challenging topic for me so far, but now I love it so much <3

Some warmup questions out there for thought (And yes there's like 3 putnam and 2 IMO's questions coming soon , this ought to be fun)

1. Show that if then

2. If show that one of the three must be a multiple of

3. Find all prime such that is a prime

4. Prove there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes

[kamil9876]

We have to show that either a or b is a multiple of 3. Assume that neither a nor b was a multiple of 3. This means:





But no square is congruent to 2 mod 3. Since

same trick for all other q's.

[TrueTears]

How do you know if a or b wasn't a multiple of 3 then it'd be congruent to 1 mod 3?

[kamil9876]

0,1,2 or equivalently 0,1,-1 are all the residues mod 3. Each integer has exactly one of these, if not a multiple of 3, then not 0 and hence 1 or -1.

[TrueTears]

0,1,2 or equivalently 0,1,-1 are all the residues mod 3. Each integer has exactly one of these, if not a multiple of 3, then not 0 and hence 1 or -1.

lol how stupid of me, how did I not think of that ffs

[TrueTears]

We have to show that either a or b is a multiple of 3. Assume that neither a nor b was a multiple of 3. This means:





But no square is congruent to 2 mod 3. Since

But



So there is a value of c that satisfies, since the question never stated a b and c are integers?

Nvm

[TrueTears]

2. If show that one of the three must be a multiple of

The questions wants us to show that

Let's assume the contrapositive, that does not divide









Now assume some number that is not congruent to satisfies the equation.

Now





But clearly this is false.

So there are no numbers that is not congruent to which satisfies the equation. So at least one of them must be divisible by .

[kamil9876]

Mods doing mod  

[kamil9876]

hint for 3:

3^2+2=11

so x=3 is a solution.

5^2+2=27=3(9) so 5 is not a soution.

7^2+2=51=3(14) so 7 is not a solution.

11^2+2=123=3(41) so 11 is not a solution.

13^2+2=171=3(57) so 13 is not a solution.

Now conjecture and prove.

[TrueTears]

3. Find all prime such that is a prime

Thanks kamil for hint.

I conjecture that the final answer can not be a multiple of and that any prime number larger than when subbed into the expression produces an integer that is a multiple of .

Assume that the final answer is a multiple of .











But all prime numbers (except for ) can be written in the form of either or .





Thus any primes in the form of or when subbed into the expression produces an integer that is a multiple of which does not satisfy the condition.

However the only prime which can not be expressed in the form of or is .




Or how about this way...

Assume that is prime iff is prime.

Since from the experimenting we conjecture that any prime larger than produces an integer that is a multiple of . Let us consider .

Since is prime, it can only be divided by itself or .







We require in order to maintain as a prime.

for to be prime.

However the only prime number in the form of is .

Thus .