One fairly easy diophantine equation anyone can solve | mathematica

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[guisav]

Nice.
Now stop the video at 1:16.
You can see from the second line that m^2+n^2+mn=0. Since squared numbers are either zero or positive, the you have two options: either m=n=0, or mn is negative. But from the first line (m+n)^2=mn, mn cannot be negative. So m=n=0 and therefore x=-1 and y=1.

[shrayanpramanik]

Another approach to the above question:
A slightly manipulated version of the equation,
(x+y)=((x+1)(y-1))½ somehow reminds me of the am gm inequality.
Point to be noted is, either both (x-1) and (y+1) is negative or both are positive.
So, considering them to be positive,
We get (x+y)²>=4(x+1)(y-1)
And if both are negative,
We can still say that,

Which gives us,
(x+y)²>=4(x+1)(y-1)
Thus, (x+1)(y-1)>=4(x+1)(y-1)
Thus, (x+1)(y-1)<=0 but it can't be negative. Thus, it's zero. Which gives us, x=-1 and y=1.

[ivankaznacheyeu]

This problem is not diophantine equation, it has the only solution in reals not only in integers

[michaelpovolotskyi]

I considered the problem as quadratic equation for x, then everything is straightforward because the Discriminant will be -3(y-1)^2

[ganesansubramaniyam]

I just equalled that (x+y)(x+y) = (x+1)(y-1) and x+y = x+1 therefore y = 1 and x+y = y-1 therefore x = -1 that's it . Awesome video

[prabhudasmandal]

m=0, n=0, understood but when (m+n) square =o, m is equal to _n and m+n = 0 becomes x+1 +y_1=0 or x+y=0 means there may be other roots too.

[screamman]

Bro because of your help im getting good at math please do another work i can't wait

[prabhudasmandal]

Tried again. If m square+mn+n square=0 multiplied by (m_n) both sides, it becomes m cube+n cube=0, i.e m=_n=x+y=0, then diophantine.

[sevenkaz]

Ok sorry firstly i thought that author is a troll then i understood that it is equation. Very interesting! Viewers remember it is not equation it is equality. My mather language Russian, i dont understand that author told.

[Siddhartha_G]

(x+y)(x+y)=(x+1)(y-1), now it is lot more easy to solve my friend

[sevenkaz]

People, wtf? (x+y)²=x²+2xy+b². I dont understand that author tell.

[Alima]

C'est vraiment la seule solution possible

[haroldlake]

Otra solución: y=-1 ; x=1. Se verifica la ecuación original.