Positive integer solutions of $0=10x³-(2y+5)x²+(y-4)x+76$ To practise my mathematical skills, I often solve some problems in my free time.
In this case, one should find every positive integer solution $(x,y)\inℤ^+\timesℤ^+$ of $0=10x³-(2y+5)x²+(y-4)x+76$.
Now this is how far I got:


*

*Alternate form: $xy(2x-1) = 10x³-5x²-4x+76$

*Alternate form: $4(x-19) = x(2x-1)(5x-y)$

*Simple solution: $(x;y)=(1;77)$ (does not make things easier regarding a factorization over $x-1$_(?)_)


But i keep being stuck with these insights… any tips?
By the way: I do not have any further knowledge in number theory and diophantine equations!
 A: Hint: From the original equation it follows that $x$ has to be a divisor of $76$. There aren't that much divisors of $76$, so let us list them down:
The divisors of $76$ are $\pm1,\pm2,\pm4,\pm19,\pm38,\pm76$. Now it's a good idea to turn back to the factorized form you discovered: $4(x-19)=x(2x-1)(5x-y)$.
A: $10x³-(2y+5)x²+(y-4)x+76 = 0$
Solving for y, we get
$y = \dfrac{10x^3 - 5x^2 - 4x + 76}{2x^2 - x}
= 5x + \dfrac{148}{2x-1} - \dfrac{76}{x}$
The only values of x that are divisors of 76 and make $\dfrac{148}{2x-1}$ an integer are $x \in \{1, 19\}$

So {$(x,y) = \{(1,77),(19,95)\}$}

It's still possible that $\dfrac{148}{2x-1}$ and  $\dfrac{76}{x}$ could both evaluate to fractions and their difference is an integer.
Say $\dfrac{148}{2x-1} - \dfrac{76}{x} = n$. This leads to 
$2n x^2 - (n-4)x - 76 = 0$, whose discriminant is 
$(n+300)^2 - 89984$ which must be a perfect square. The only values of n that make it a perfect square and make $x$ an integer are $n=0$ and $n = 72$. Which gives us the values of x that we have already found.
A: since $68590-70395+1729+76=0$ hence $(x-19)$ is definitely a factor of this polynomial so we must frame this equation in this particular manner:
$$10x^3-195x^2+91x+76=10x^3-190x^2-5x^2+95x-4x^2+76=(x-19)(10x^2-5x-4)$$
