How to prove that there’s no positive integers $x,y>3$ such that $(x^2-2)y^2 -2xy-(2x^2-2)=0$
#quadratic
plucky hinge
young shoreBOT
errant oysterBOT
Low
plucky hinge
we can use b^2 - 4ac right
torn sentinel
what class / math topic is this question from?
is this intro to proofs? number theory? olympiad problem?
or some kind of calc class
plucky hinge
is this intro to proofs? number theory? olympiad problem?
oly problem
i guess
i saw it on a server
torn sentinel
yeah thought so, had that kind of vibe
plucky hinge
ok
plucky hinge
**Low**
maybe we can prove that it’s >0
by using some asymptomatic ways
What do you think
torn sentinel
doubt it, since the question is working with a hard bound on x,y
I can prove asymptotically that there must exist a big enough N such that x,y>N has no solutions, that's not too hard
the difficulty here is working with the given condition x,y>3
see below
plucky hinge
Ok
plucky hinge
~~I can prove asymptotically that there must exist a big enough N such that x,y>...
Ok but we need to prove for N = 3 , right
torn sentinel
by using some asymptomatic ways
so your idea was actually pretty good, because the realisation hit me that as long as we can find a numerical value for N, even if N is greater than 3 we can just manually check all the smaller values as long as N is small enough. So asymptotic approaches do work
Suppose $x+y = N$.
Then as $x,y \geq 4$ by assumption, we have that $xy \geq 4(N-4)$.
Now rearrange the equation to obtain $2(x+y)^2 = (xy+1)^2+1$. Then:
\begin{align*}
[(4N-16)+1]^2+1
&\leq 2N^2 \
&\vdots \
14N^2-120N+1 &\leq 0 \
N &\leq 5.78
\end{align*}
But $N \geq 8$, which is a contradiction. $\blacksquare$
errant oysterBOT
HChan
torn sentinel
damn this was a fun question.
also fuck you I had work to do that was 2 hours of my life I'm not getting back
plucky hinge
Thank you!
torn sentinel
Where did you get this question from?
plucky hinge
@torn sentinel maybe i need some explanations
first xy=> 4(N-4) by am gm, right
Then, i don’t understand how do u get ((4N-16) + 1)^2 + 1 =< 2N^2
could u give more details
plucky hinge
first xy=> 4(N-4) by am gm, right
No
plucky hinge
@torn sentinel i just get the problem
Find every positive integers such that $(x^2 - 1)(y^2 - 1) = (x + y)^2 - 1$
errant oysterBOT
Low
plucky hinge
The initial problem
torn sentinel
Find every positive integers such that $(x^2 - 1)(y^2 - 1) = (x + y)^2 - 1$
the method I used to solve the other problem can be repeated here
the only solutions in the positive integers are x=2,y=3 / x=3,y=2
fun problem
plucky hinge
@torn sentinel
torn sentinel
It is am gm, just a slightly different version of it
plucky hinge
could u prove it ?
torn sentinel
two ways to prove this, either with calculus or with number theory:
First fix a positive integer N.
Then if x+y=N, xy=x(N-x).
Furthermore, we can WLOG assume that x <= y, so we in fact only need to consider x(N-x) for values of x between 0 and N/2. Furthermore as x>3, we can further restrict the domain to 4 <= x <= N/2.
Proof 1 (number theory):
AM/GM tells us that the number xy is maximised when x is close to y. Conversely, this means that xy is minimised when x is the furthest away from y, and since y has to be between N/2 and N, the furtherest away from y that x can get is x=4. Thus the minimum of xy is achieved when x=4 and y=N-4, done.
Proof 2 (calculus):
The quadratic equation x(N-x) has roots at 0 and N, so its global maximum is at x=N/2. But then the quadratic is increasing for 4 <= x <= N/2, and so it follows that the global minimum of the quadratic is found on the endpoints. Thus the minimum is at x=4, y=N-4, done.
plucky hinge
Ok i see
torn sentinel
direct substitution
torn sentinel
where do you get these from?