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the question

Determine the pairs $(x,y)$ of integers with the propriety that $$2x^2-3xy+y+1=0$$

my idea

I tried writing it as a product of terms but got to nothing useful. Then I applied the quadric formula and got that $x_{1,2}=\frac{3y+-\sqrt{9y^2-8y-8}}{4}$ and $x,y$ must be integers which will make $\sqrt{9y^2-8y-8}$ be a rational number so $9y^2-8y-8$ is a square number.

I don't know what to do forward! Hope one of you can help me!

nonuser
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IONELA BUCIU
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    Hint: Try to put $9y^2-8y-8$ between two square numbers. – Zumzumike Apr 11 '24 at 18:37
  • Does it help to note that $y(1-3x)$ must be odd? Thus, $y$ is odd and $x$ is even. Now think mod 8. – Ted Shifrin Apr 11 '24 at 18:45
  • $81 y^2 - 72 y - 72 = (9y-4)^2 - 88$ – Will Jagy Apr 11 '24 at 19:07
  • Same method as in here in the dupe, e.g. see my comments here for a very similar problem, viz. $,g=\color{#c00}{3x!-!1}\mid 2x^2!+!1!\iff! g\mid 3^2(2x^2!+!1)=2(\color{#c00}{3x})^2!+!9!\iff! g\mid 2(\color{#c00}1)^2!+!9,,$ by $,\color{#c00}{3x\equiv 1}\pmod{!g}\ \ $ – Bill Dubuque Apr 13 '24 at 06:58
  • Or, in gcd language (as in linked dupe) $\ g = (g,\color{#0af}{2x^2!+!1}) = (g,\color{#0af}{11})\iff g\mid 11,,$ using, as above, $,\color{darkorange}3^2(\color{#0af}{2x^2!+!1})\equiv \color{#0af}{11}\pmod{!g},,$ and $,1=(g,\color{darkorange}3)=(3x!-!1,3)\ \ $ – Bill Dubuque Apr 13 '24 at 07:25
  • Duplicate, see above links. Should not have been reopened. – Bill Dubuque Apr 13 '24 at 15:35

3 Answers3

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Hint 1:

$$2x^2+1=(3x-1)y\implies 3x-1\mid 2x^2+1$$

Hint 2:

$$3x-1\mid 3(2x^2+1)-(3x-1)2x = 2x+3$$

Hint 3:

$$3x-1\mid 3(2x+3)-2(3x-1)= 11$$ So $3x-1\in \{\pm 1,\pm 11\}$ so $x =0$ or $x= 4$...

nonuser
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Completing squares we have $$(12x-9y)^2-(9y-4)^2+88=0$$ which yields to $$(A+B)(B-A)=88=2^3\times11$$ where $A=12x-9y$ and $B=9y-4$

Among the two factors giving the product $88$ we need both even so we discard $8$ with $11$ and $1$ with $88$ so remains the only possibilities $$A+B=22\\B-A=4 $$ and $$A+B=44\\B-A=2 $$ these systems have the solutions $$(A,B)=(9,13)\text { and }(21,23)$$ which yields the systems $$\begin{cases}9y-4=13\\12x-9y=9\end{cases}\text { and } \begin{cases}9y-4=23\\12x-9y=21\end{cases}$$ Only the second has integer solution $(4,3)$.

There is just one solution: $\boxed{(x,y)=(4,3)}$

Ataulfo
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0

Since $y+1\equiv0\pmod x$, substituting $y=-1+kx$ in the equation we get $x=\frac{k+3}{3k-2}$. Hence $x=4$ or $x=0$.

Bob Dobbs
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