About the Rules & Moderation category (Part 1)

Announcements Rules & Moderation
#13559

read slightly below that

#13560

that’s a non sequitur

#13561

my solution was

solution

there are infinite perfect squares m

let n = (m^2) / 4 + m
n = (m/2)^2 + (sqrt(m)) ^ 2
n + 1 = (m/2 + 1)^2 + 0^2
n + 2 = (m/2 + 1)^2 + 1^2

#13562

thats irrelevant

#13563

yeah i knew all that

1 Like
#13564

i do think you could do it with a number theory argument. every number that’s 0, 1, or 2 mod 4 can be written as the sum of two perfect squares i think. it’s impossible for any number that’s 3 mod 4 to be the sum of two perfect squares at the very least so n has to be 0 mod 4

#13565

12 isn’t the sum of two squares

#13566

okay this seems false idt you can write 6 as the sum of two perfect squares. not every case. but the argument’s there

#13567

n has to be 0 mod 4. you can do something with that

#13568

number theory rules all

#13569

there are infinitely many such cases

#13570

Very sad! Infinitely many such cases!

#13571

give me 3 hours at this and i can write down a vaguely plausible sounding solution

#13572

nyaaaaaaaaaaaa open my spoiler :flushed:

#13573

i did. it’s not number theory. unfun

1 Like
#13574

oh I guess I rely on m being divisible by 2 as well (which is fine tho)

#13575

well it is. but it’s not the number theory i was thinking of. i do think you could do it that way

#13576

NYAAAAAAAAAAAAAAAAA I haven’t taken number theory (sad)

#13577

well every perfect square is either 0 or 1 mod 4. you can use that (that also trivially explains why 3 mod 4 can never be the sum of two perfect squares)

1 Like
#13578

i love modular arithmetic

1 Like