Turkey NMO 1997 Problem 1, Diophant Equation

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Turkey NMO 1997 Problem 1, Diophant Equation

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Source: Turkey NMO 1997 Problem 1

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#1 h Sep 28, 2011, 8:42 AM • 3 Y

Y by ahmedosama, Adventure10, Mango247

Find all pairs of integers $(x, y)$ such that $5x^{2}-6xy+7y^{2}=383$ .

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#2 h Sep 28, 2011, 9:47 AM • 2 Y

Y by Adventure10, Mango247

I think there are no solution for this problem.

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#3 h Sep 28, 2011, 10:18 AM • 3 Y

Y by pankajsinha, Adventure10, Mango247

There are: $(10, 3)$ and $(-10, -3)$ .

Solution. Rewrite the equation into $(x-y)^2+(2x-y)^2+5y^2=383$ . So $y^2 \leq \frac{383}{5} \implies |y| \leq 8 \implies y \in [-8,8]$ . Now, rewrite the equation as $5x^2-6xy+7y^2-383=0$ and solve the quadratic equation for $x$ . The discriminant is $\Delta=7660-104y^2$ , and it should be a perfect square. So, suppose that $a^2=7660-104y^2$ . Use $\mod 3$ and you get $y \equiv 0 \pmod 3$ . But we had $y \in [-8,8]$ , so $y=-6,-3,3$ , or $6$ . You can easily check the answers and see that $y=3$ and $y=-3$ satisfy the equation.

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#4 h Sep 28, 2011, 10:31 AM • 4 Y

Y by abhraabirk, pankajsinha, Adventure10, Mango247

multiplying with 5 gives us $(5x-3y)^2+26y^2=1915$ then try $y=1,2,\cdots8$ gives us solution : $(10,3),(-10,-3)$

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#5 h Oct 14, 2015, 2:42 AM • 3 Y

Y by hodori01, Adventure10, Mango247

解: 原方程配方, 得
$(5x-3y)^2+26y^2=1915. \qquad (1)$ 因此
$26y^2\leqslant 1915.$ $y^2\leqslant 73\frac{17}{26}. \qquad (2)$
另一方面, 由 $(1)$ 式, 知 $5x-3y$ 是奇数, 故对 $(1)$ 式模8, 得
$1+2y^2\equiv 3\pmod{8}.$ 而 $y^2\equiv 0, 1, 4(\bmod{8})$ , 仅有 $y^2\equiv 1(\bmod{8})$ 满足上式. 从而 $y$ 是奇数.

对原方程式模3, 得
$-x^2+y^2\equiv 2\pmod{3}.$ 而完全平方数模3余数只能是0或1, 所以
$\begin{aligned} x^2&\equiv 1\pmod{3}, \\ y^2&\equiv 0\pmod{3}. \end{aligned}$ 故 $y$ 是3的倍数. 结合 $(2)$ 式(注意 $y$ 是奇数), 知
$y=\pm 3.$
所以
$\begin{cases} y=3, \\ (5x-9)^2=1915-26\times 3^2=41^2; \end{cases}$ 或
$\begin{cases} y=-3, \\ (5x+9)^2=1915-26\times 3^2=41^2. \end{cases}$
由模5, 可知
$\begin{cases} y=3, \\ 5x-9=41; \end{cases}$ 或
$\begin{cases} y=-3, \\ 5x+9=-41. \end{cases}$ 解得
$(x, y)=(10, 3), (-10, -3).$

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#6 h Apr 14, 2025, 8:26 PM

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$(x-y)^2+(2x-y)^2+5y^2=383$ . So $y^2 \leq \frac{383}{5} \implies |y| \leq 8 \implies y \in [-8,8] \implies 5x^2-6xy+7y^2-383=0$ $x$ . The discriminant is $\Delta=7660-104y^2$ $a^2=7660-104y^2$ . Use $\mod 3$ and you get $y \equiv 0 \pmod 3$ . But we had $y \in [-8,8]$ , so $y=-6,-3,3$ , or $6$ . You can easily check the answers and see that $y=3$ and $y=-3$ satisfy the equation.

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