Solving 117y^3-x^3=5, a Diophantine Equation

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[SyberMath]

Apparently I made two videos on Modular Arithmetic! 😂

[arundhatimukherjee]

117y³-5=x³ or 117y³-9+4=x³, so x³ is 4 mod 9. But cube of an integer is either 0, 1 or 8 mod 9. So no integer solutions🙂🙂

[yoav]

I knew syber will give us more than 1 method😀

[robertgerbicz]

You can save that large table at the end, because from Euler Fermat if gcd(x, 3)=1 then x^6==1 mod 9, so
9|x^6-1=(x^3-1)*(x^3+1) but then
x^3==+-1 mod 9 (and ofcourse x^3==0 mod 9 if 3 divides x).

[ابوسالم-هم]

We get

And this is satisfy only when y= 1& x=5 which are did not satisfy original equ

[yoav]

This is the most hated solution: no solution.😄😁

[kylekatarn]

So, when you change the mod you set a number equal to its rest. I do not understand what do you do in the last transformation. For example in the second method you set x^3=-5 (mod 9) and then you put x^3=4 (Mod 9), how is that?

[Qermaq]

Re: x^3 - 117y^3 = 5 -- I checked all values of x from -25000 to 25000 and none produce an integer y. What was the equation you meant at the end?

[ridazouga]

Well I solved it just under 5 minutes, because if this equation had a solution, then we can prove that GCD(x, 9)=1, then by using Euler's totient function(wich is a generalization of Fermat's little theorem) that x^6=1(mod 9), but if we go back now to our equation we get x^3=4(mod 9) and thus x^6=7(mod 9) and thus CONTRADICTION!

[spencergee]

I'm just making dinner, otherwise I would solve it!

[WolfgangKais]

Your last equation has no solution, because if (x, y) was a solution, then (-x, -y) would solve the original problem.

[yoav]

But i wonder what will be the method of syber and how many

[prabhudasmandal]

When there is no solution, why should we call it an equation?

[moeberry]

For the last equation you gave at the very end of the video, there is NO solution because x^3 is never congruent to 5 mod(9) but yet you stated there’s at most 18 solutions.