#Im completely stuck
35 messages · Page 1 of 1 (latest)
can you factor x^2 - y^2
Oh ok so $(x+y)(x-y) = 1, (x+y) = (x-y)$
$2y = 0, y = 0$
$x^2 = 1$
Kardiiacc
am i on the right line?
how are you going from (x+y)(x-y) = 1 to (x + y) = (x-y)?
you are certainly going to reason about x and y from the first step that you did
but the right line here is just to think about what it means for two integers to multiply to 1
Yo my apologies i literally made an error when calculating this ima do it again soz
it's okay!
but you are done with the symbolic manipulation once you factor in the first step
you just need to realize that it's not possible for (x+y)(x-y) to ever be 1 (why is this true?)
(remember that x and y are positive integers)
Bc nothing multiplies together to get 1 except 1 and 1
exactly (or -1 and -1, but you will never have that case, since x and y are positive)
so why can't both factors be 1
what if i told you that both factors were 1 what would you say to me
x would have to be negative as its both -y and +y to get 1
or 0*
Oh wait i think i got that wrong
there is a clearer reason
think about what we supposed for the sake of contradition - that x and y were both positive integers
So (x+y) = 1 and (x-y) = 1 and by sake of contradiction, there is atleast on positive integer that has the solution x^2-y^2 = 1. Use simulataneous equation to solve for y. 2y = 0, y = 0. Plug in y and x = 1. but there is atleast one positive integer tho
Im still a bit stuck
well what would it mean for x + y = 1
specifically
we can get a contradiction right here
Ohhh there is no positive integer that can add to get 1
yep 
Lolll i didnt even realise
lol i remember doing this problem for the first time and it was really confusing to me
and then i saw it and I was like omg
okay so you should write out the full contradiction proof (either on your paper or here) and then you're done
are you comfortable with writing a contradiction proof?