Prove that every odd integer is the difference of two squares
i suppose best proof to use would be contradiction...
so i assume x^2 - y^2 = 2k
then... for x^2 - y^2 to be even... both must be even or odd
so if x^2 and y^2 are both even... x and y are also even
no we're trying to prove converse i think, which doesnt justify the actual statement
so i can rewrite this... (2m)^2 - (2n)^2 = 2k
4m^2 - 4n^2 = 2k
so.... 2m ^ 2 - 2n^2 = k
then... m^2 - n^2 = k/2
can i assume this is contradiction because 1/2 is not an integer?
could u plz clarify, you want to prove difference of two squares is alwasy an odd number, or you want to prove every odd number is a difference of two squares ?
no idea
how do you understand the question?
the question is, every odd integer can be written as difference of squares
but i see you're trying to prove its converse, which is not correct
so how would you do the assumption?
using contradiction
personally, i see no difference in solution
consider below two statements : 1) all odd numbers are diff of squares 2) all diff of squares are odd numbers
both are converses of each other
proving by contradiction is different
so tell me...can you think of a different solution then?
what would you equate 2k + 1 to in proof by contradiction then?
if you think about it...every difference of two squares is also odd...
first, did u get that a converse is not same as the actual statement proving a converse doesnot justify the actual statement
i know...but i cannot see any other way to prove by contradiction....so if you have an idea...i'll gladly hear it
i think , it means : (2n-1) = n^2 - (n-1)^2 just to expand right side, it can be a prove
@RadEn contradiction
then dont try to prove the converse, lets think of how to prove by contradiction, by assuming the opposite of what we need to prove :)
what is the opposite of difference of two squares? sum??
we can factor difference of squares
then you're using direct proof
doesnt look obvious to me the contradiction proof
looks involved
involved?
yes atleast to meh il give a try
i think you'll arrive with the same proof i was doing earlier...
thats not a proof
it is actually...probably not contradiction...but it is
it is a proof*
thats not a proof for this question, it is a proof for the converse of this question
you're welcome to give it a try
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