Can You Solve This Diophantine Equation? | Romanian Math Olympiad | Number Theory

Can you solve this number theory problem from Romania?

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Maths videos by Giuliano Grasso - mathematics graduate from the University of East Anglia.

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[farhatali]

Thanks for your effort s. Please continue the series.

[djvalentedochp]

I just completed the square with fractions instead of multiplying the equation by 4 and got the same answer, nice problem!

[wesleydeng]

x^3(x^3+3)=(y^2-1)(y^2+1). let a=x^3, b=y^2-1 => a(a+3)=b(b+2), a>=0, b>=0.
If a and b are not 0, then obviously a<b, so a<=b-1. But in this case, a(a+3)<=(b-1)(b+2)<b(b+2), contradiction. So a=b=0 => x=0, y=1, -1.

[jamescollis]

you can rule out 2 of the 4 potential solutions because y^2>=0 => 2x^3+2y^2+3>=2x^3-2y^2+3 and (5, 1), (-1, -5) only potential solutions

[prabhudasmandal]

If we take x cube=a, the equn becomes a square+3a+(1_y square)=0, a quadratic where a is 1, b is 3 and(1_y square) is c and solving finally we get y=+/_1 and x=0.

[glasssmirror]

Hi sir Pls how in this world would you know that we have to multiply by 4 all round

[davidgillies]

I substituted x^3 -> u, y^2 -> v so u(u + 3) = (v + 1)(v - 1). I then substituted v -> w +1 so u(u + 3) = w(w + 2). The only case for which this holds true is u = w = 0 and v = 1 whence the solution.

[mcwulf]

You could probably put any polynomial of X on the left with the x^0 term = 1 and the only solutions are (0, +/-1).

[AstroB]

Hi !
Nice video !
Also, what software are you using to draw/write ? Thank you

[jarikosonen]

This looks possible solution, but is it the only solution.. maybe it is.
y=(x^6+3*x^3+1)^(1/4), then x=0, gives y=1.
Can this method find all the integer solutions for the pythagorean triangle edge lengths? (like 3, 4, 5, etc)