About the Rules & Moderation category (Part 1)

Announcements Rules & Moderation
#13859

i got the correct answer for A1 but i wrote down some utter nonsense. maybe i get partial credit?

i also think i got B1 possibly correct but i haven’t discussed that one. and also didn’t justify things super greatly

1 Like
#13860

yeah we also did. there was one person who managed to prove it didn’t work for n = 2… during lunch. after they had already turned in that question

1 Like
#13861

i hope they at least appreciate my “attempt” on B…5? i think? whichever one thae “prove it’s a polynomial” one was. the putnam graders need more whimsy in their life

#13862

i should’ve committed to my lunchtime promise to just start writing fanfiction on one of the problems. but i wanted to leave early

#13863

my solution was basically:

for n>2:

2a^n + 3b^n = 4c^n
→ b congruent to 0 mod 2

2a^n + 3b^n = 4c^n
→ 2a^n + 3(2k)^n = 4c^n
→ a congruent to 0 mod 2

2a^n + 3b^n = 4c^n
→ 2(2j)^n + 3(2k)^n = 4c^n
→ c congruent to 0 mod 2

Because you can get all these, you can divide them all to get solutions that aren’t 0 mod 2, which is a contradiction

then for n=2, you do:
2a^2 + 3b^2 = 4c^2
→ -a^2 = c^2 (mod 3)
→ a = c = 0 (mod 3)

then
2a^2 + 3b^2 = 4c^2
→ 2(3k)^2 + 3b^2 = 4(3j)^2
→ b = 0 (mod 3)

then the same idea of dividing all of them to get a contradiction

1 Like
#13864

somebody asked chatgpt to solve A1 for the bit. it used the most purple prose i’ve ever seen in a math proof. or seen chatgpt use ever. it described the equation as “crystalline” at one point

3 Likes
#13865

they did that for n > 2 but couldn’t figure out n = 2 until lunch. if you do things mod 8 you can get a contradiction though

1 Like
#13866

it gave up after n = 1

3 Likes
#13867

LOL

1 Like
#13868

it did manage to verify that a = 1, b = 1, c = 1 doesn’t work for n = 2. i’m proud of it!

2 Likes
#13869

Putnam moment

1 Like
#13870

unfortunately i did not have the “you can always find a lesser solution” brainwave. because i’m cooked. i started writing some nonsense about the sum of odd numbers formula for perfect squares and tried to do induction from there

2 Likes
#13871

i also spent the vast majority of the allotted 3 hours on A1. literally got nerd sniped. as i tried to explain during lunch

1 Like
#13872

i think B1 was actually easier than A1. or at least i figured out what the answer probably was easier. idk if i did a good job actually showing it but the vision was there

#13873

btw A3 being yes is implied by an unproven conjecture! I was VERY tempted to just HOPE the conjecture was proven in the past months of me not looking at it

1 Like
#13874

which one was A3 again

#13875

i “submitted” something for A2 that was just me giving up after 2 sentences.

and i submitted a not very well explained answer of “none” for the sequences mod primes one in the remaining 15 minutes i had

#13876

let S be the set of bijective functions T where

T : {1, 2, 3} X {1, 2, 3 … 2024} → {1, 2, 3, … 6,072}

And the following hold:
T(i, j) < T(i, j + 1)
T(i, j) < T(i + 1, j)

(for the ranges that make those defined of course)

Does there exist a,b,c,d such that the proportion of T in S where T(a,b) < T(c, d) is at least 1/3 and at most 2/3?

#13877

answer is implied to be yes by this conjecture

#13878

right that one. yeah i don’t think anyone here actually tried it

1 Like