$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$

Question

suppose $n\in \Bbb Z $ then how to prove this statement:

http://math.stackexchange.com/questions/263101/prove-every-odd-integer-is-the-difference-of-two-squares will be of interest. Also http://math.stackexchange.com/questions/2462/multiples-of-4-as-sum-or-difference-of-2-squares — Gerry Myerson, Mar 12 '13 at 06:55

agustin · Accepted Answer · 2013-03-12T07:26:16.323

2

tnx for Gerry Myerson. it's equal with this statement:

$x^2-y^2=n$ iff $ n$ is odd or $4|n$

$(x-y)(x+y)=x^2-y^2=n=ab , \exists a,b \in \Bbb Z$

so we have ($x+y=n$,$x-y=1$ or $x-y=n$,$x+y=1$ ) that $n$ must be odd or $x=\frac{a+b}{2}, y=\frac{a-b}{2}$ that so two case exists :

$I)$both $a,b$ is even so $4|ab=n$

$II)$ both $a,b$ is odd so $n=ab $ is odd

edited Mar 12 '13 at 07:26

answered Mar 12 '13 at 07:17

agustin

score 1 · Answer 2 · answered Mar 12 '13 at 08:16

Observe that $x+y=x-y+2y$

So, $x+y,x-y$ have the parity

If they are odd, $n=x^2-y^2$ will be odd

If they are even, $n=x^2-y^2$ will be divisible by $4$

Conversely, if $n$ is odd $=2m+1$(say)

$(x+y)(x-y)=2m+1$ where $m$ is any integer

So, we can write $x+y=2m+1,x-y=1\implies x=m+1,y=m$

or $x+y=-(2m+1),x-y=-1\implies x=-(m+1),y=-m$

If $4\mid n,n=4r$(say)

So, we can write $x+y=2r,x-y=2\implies x=r+1,y=r-1$

or $x+y=-2r,x-y=-2\implies x=-(r+1),y=-(r-1)$

2 Answers2