British Math Olympiad 2012 Problem | Number Theory

British Math Olympiad 2012 Problem | Number Theory

[echandler]

mod 4 is the remainder on dividing by 4 or equivalents. Thus you get {0, 1, 2, 3} or these plus or minus any multiple of 4, hence {0, 1, 2, -1} is equivalent. squaring these gives {0, 1, 0, 1}. Reducing mod 4 shows that the exponent of 3 must be even. Then you can form 8 as the difference between two squares, which factors. 8 can be decomposed into factor pairs in a limited number of ways.

Reducing mod (something), factoring the difference of two squares and unique factorization are common techniques in solving elementary number theory problems.

[tiernanomahoney]

This is exceedingly elementary for BMO, surely it's a lot harder than this

[thichhochoi]

m must be odd. 3^k-1 and 3^k+m are even, no need to check for case (1, 8)

[navaneethvijay]

Can you explain me the mode form in this problem ?
I need to know about it

[xiaoshen]

Wtf ye bhi intuition se hi ho gya 😭. Ye bahut galat aadat h (guessing value and putting wali )

[jonathanrajan]

nice video!
which bmo paper was this in? I checked 2012 and it's not there

[pomlesty]

I have a really weird proof of it
So let's begin.
m²+8=3^n
8=(3^n/2)^2-m²
8=(3^n/2 + m)(3^n/2 – m)
Here I assumed that both factors of 8 are integers
But it can be proved that both factors are positive

[geraldturner]

What the heck is mode 4???
What is mode BTW?