Question

A simple algebra challenge

\[\large{\color{red}{\text{THINK OUT OF THE BOX!!}}}\]

Answer 1

Let r and s be non zero integers. Prove that the equation

\[\large{(r^2-s^2)x^2 - 4rsxy - (r^2-s^2)y^2 = 1}\]

has no solutions in integers x and y

Answer 2

@Kainui @ganeshie8 @JungHyunRan @Miracrown @hartnn

Answer 3

Would r and s represent trigonometric values?

Answer 4

Or by simple algebra?

Answer 5

You have all the things given in the problem. You can use whatever you want but don't forget to justify your any assumption.

I would prefer using simple algebra

Answer 6

\[s = r \theta \]

Answer 7

Alright thank you.

Answer 8

r and s are integers - remember this line :)

Answer 9

b^2 -4ac
= [16 r^2 s^2 +4 (r^2-s^2)^2 ]

= 4[4 r^2 s^2 +r^4 -2r^2s^2 +s^4 ]
= 4 [r^4 -2r^2s^2 +s^4]
=4 (r^2-s^2)^2

which is a perfect square

Answer 10

(4rs +/- 2(r^2-s^2))/2 (r^2-s^2)

= -(r+/-s)^2 / (r^2-s^2)

= -(r+s)/(r-s) or -(r-s)/(r+s)

Answer 11

i guess thats thinking inside the box :P

Answer 12

Great but how will you prove the statement using this ?

Answer 13

Okay let me provide my solution...

Answer 14

We suppose that x and y are integers satisfying the given equation. Factoring the LHS, we get:

Answer 15

\[\large{((r-s)x - (r+s)y)((r+s)x + (r-s)y) = 1}\tag{1}\]

Answer 16

wow! did anyone realize the blunder i made :P

Answer 17

Yeah +4 -2 = -2

Answer 18

I'm still trying to figure it out actually haha.

Answer 19

I did this in my head just can't type it, my hands are broken

Answer 20

lol , i took quadratic formula!
applicable only for
ax^2+bx+c = \(\huge 0\)
not for
ax^2+bx+c = 1
:P

Answer 21

Hahaha

Answer 22

Okay let me continue:

as each factor in LHS of equation (1) is an integer, we can obtain the following:

\[\large{(r-s)x - (r+s)y = \epsilon} \tag{2a}\]
\[\large{(r+s)x + (r-s)y = \epsilon}\tag{2b}\]

Where, \[\large{\epsilon = \pm 1}\]

Answer 23

Solving equations (2a) and (2b), I get:

\[\large{x = \cfrac{r\epsilon}{r^2+s^2}}\]

\[\large{y = \cfrac{-s\epsilon}{r^2+s^2}}\]

Hence, we obtain :

\[\large{(x^2+y^2)(r^2+s^2) = 1}\]
So that:

\[\large{r^2+s^2 =1}\]

\[\large{\implies (r,s) = (\pm 1, 0)~or~(0, \pm 1)}\]

which is not possible as r and s are both non zero.

Thus, given equation has no integral solutions.

Answer 24

\[\huge{\color{blue}{\text{Q.E.D.}}~~\color{red}{\text{Vishwesh}}}\]

Answer 25

Well done @hartnn . Nicely tried :)

Answer 26

A mathematician who does not know to show off his work is an incomplete mathematician.

That is why I used \(\large{\epsilon}\) in place of some other simple symbol like "k".