Question

Number theory

Answer 1

the question:

n is an integer and we have \(x^2 - y^2 = n\),prove that this equation is true for integer numbers if and only if \( n \not\equiv 2 ~ (mod ~ 4~) \)

Answer 2

my proof:

Answer 3

\( \huge \color{blue}{(1)} \) \(x^2 , y^2 \equiv 0 ~ or ~ 1 ~ (mod ~ 4) \) so \( n\equiv 1 ~ or ~ 0 ~ or ~ 3 ~ (mod ~ 4) \)


\( \huge \color{blue}{(2)} \) \( n\not\equiv2 ~ (mod ~4)\) so, \( n\equiv0 ~ or ~ 1 ~or ~ 3 ~ (mod ~4)\) so, n can be written like this:

\( \large n=4k ~ \color{red}{(1)}\)
\( \large n=4k + 1 ~ \color{red}{(2)}\)
\( \large n=4k + 3 ~ \color{red}{(3)}\)

but for all of them we can say that whether x or y is odd or even and then solve the question...for example for the second one:

n is odd,so we can consider x to be odd and y to be even,let \( x=2s+1 \) and \( y=2t\).


\[x^2 - y^2 = (2s+1)^2 - (2t)^2 = 4s^2 + 1 + 4s - 4t^2 \]

so,\(x^2 - y^2 = 4(s^2 + s - t^2 ) + 1\)
finally,let \( s^2 + s - t^2 = k \).so n can be written in that way.

for all of them we can do the same things,but for first one x,y can be both odd or even and we can consider both cases and solve the question,and for the third one we consider x to be even and y to be odd and it would be solved.

Answer 4

is it true ?

Answer 5

ok i dint read fully but i'll post mine
note that , this is a theorem to solve quadratic Diophantine equation

Answer 6

@Marki i'll be glad to have another proof ;)

Answer 7

so this how i started :-
\(x^2-y^2=n \iff n\neq 2 \mod 4 \)

\(\rightarrow \) let \(x^2-y^2=n \)

\((x-y)(x+y)=n \)

16 cases xD
x=0 mod 4 , y=0 mod 4 , n=0 mod 4
x=0 mod 4 , y=1 mod 4 , n=1 mod 4
x=0 mod 4 , y=2 mod 4 , n=0 mod 4
x=0 mod 4 , y=-1 mod 4 , n=-1 mod 4


x=1 mod 4 , y=0 mod 4 , n=1 mod 4
x=1 mod 4 , y=1 mod 4 , n=0 mod 4
x=1 mod 4 , y=2 mod 4 , n=1 mod 4
x=1 mod 4 , y=-1 mod 4 , n=0 mod 4

x=2 mod 4 , y=0 mod 4 , n=0 mod 4
x=2 mod 4 , y=1 mod 4 , n=-1 mod 4
x=2 mod 4 , y=2 mod 4 , n=0 mod 4
x=2 mod 4 , y=-1 mod 4 , n=-1 mod 4

x=-1 mod 4 , y=0 mod 4 , n=-1 mod 4
x=-1 mod 4 , y=1 mod 4 , n=0 mod 4
x=-1 mod 4 , y=2 mod 4 , n=1 mod 4
x=-1 mod 4 , y=-1 mod 4 , n=0 mod 4

Answer 8

now
\(\leftarrow \) lets assume n=2 mod 4 , ( check cases before )
by contradiction , there is no such x,y integers such that
(x^2-y^2)mod 4=2

Answer 9

does this make sense ?

Answer 10

nice :)

Answer 11

remember this is 2 way proof direction :)
my first comments proof that for x^2-y^2 integers \( n\neq 2 \mod 4\)
second comment for \( n= 2 \mod 4\) there is no x,y s.t x^2-y^2=2 mod 4

Answer 12

@ganeshie8 , thank you
@Marki , thank you for your proof...i'll note your idea in my book too :) but one thing...your second one did not make sense.you used the question in your second comment??!

Answer 13

haha ,actully its only logic trick ;)

Answer 14

see 16 cases provided in general what x,y can be
so i assumed n=2 mod 4 , check cases no x,y makes it true
thus we say this is a contradiction

Answer 15

I think his proof also works as he is using the fact that any perfect square is 0 or 1 mod 4

so only two cases will do right ? @Marki

Answer 16

yes !!!

Answer 17

4 cases

Answer 18

yes yes total 4 cases :
x even y even
x even y odd
x odd y even
x odd y odd

Answer 19

hmm not even odd its with respect to 4

Answer 20

Oh right!

Answer 21

x=4m
x^2=0 mod 4
x=4m+1
x^2=1 mod 4
x=4m+2
x^2= 0 mod 4
x=4m+3
x^2=1 mod 4

nice :)

Answer 22

@Marki read ur comments carefully and i got it,nice proof :) in another words,we proved that if that equation is true,then n is not congruent to 2 (mod 4) , so if n is congruent to 2 (mod 4) then there would be no answer for that equation,therefore for \(n \equiv 2 (mod4) \) we have answer...