Positive integer solutions of $0=10x³-(2y+5)x²+(y-4)x+76$

Question

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:

1. Alternate form: $xy(2x-1) = 10x³-5x²-4x+76$
2. Alternate form: $4(x-19) = x(2x-1)(5x-y)$
3. 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!

Answer 1

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: $4(x-19)=x(2x-1)(5x-y)$.

Answer 2

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: $$10x^3-195x^2+91x+76=10x^3-190x^2-5x^2+95x-4x^2+76=(x-19)(10x^2-5x-4)$$

Answer 3

$10x³-(2y+5)x²+(y-4)x+76 = 0$

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 $x \in \{1, 19\}$

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

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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.