1989 OIM Problems/Problem 6

Problem

Prove that there's and infinity of pairs of natural numbers that satisfy the equation: $2x^2-3x=3y^2$

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

We claim that there do not exist any solutions to this equation over the positive integers.

Assume for the sake of contradiction that there exists a solution over the positive integers. First, the equation implies that $2x^2=3(x+y^2)$ ; clearly then $3|x$ , so let $x=3a$ . Then the equation, after simplification, becomes $6a^2=3a+y^2$ . Now $3|y$ from this equation, so let $y=3b$ ; therefore, $2a^2=a+3b^2$ for positive integers $a$ and $b$ .

Solving the quadratic in $a$ yields $a=\frac{1\pm\sqrt{1+24b^2}}{4}$ Clearly, for $a$ to be rational, let alone an integer, $\sqrt{1+24b^2}$ must be an integer. Let $z=\sqrt{1+24b^2}$ ; then $z^2-24b^2=1$ . This is the Pell Equation for $n=24$ . Testing values yields the minimal solution $(z_1,b_1)=(5,1)$ . (Also notice that $z^2=24b^2+1\ge24+1=25$ since $b$ is a positive integers, hence $z\ge5$ .) Then all solutions $z_k$ in $(z_k,b_k)$ can be generated by the recurrence $z_k=z_1z_{k-1}+24b_1b_{k-1}$ Taking the equation $\pmod{4}$ : $z_k\equiv z_1z_{k-1}\pmod{4}$ We showed earlier that $z_1=5$ ; hence, $z_k\equiv 5z_{k-1}\equiv z_{k-1}\pmod{4}$ for all integers $k\ge2$ . Since $z_1\equiv1\pmod{4}$ , we can use the recurrence to conclude that $z_k\equiv1\pmod{4}$ for all such $k$ . Therefore, all integer values of the square root must have this be true. Thus, $a=\frac{1\pm z_k}{4}$ But notice that the numerator must be positive, and since $z_k$ is an integer, the $\pm$ must be a $+$ , so $a=\frac{1+z_k}{4}$ However, $1+z_k\equiv2\pmod{4}$ , so the numerator is not divisible by the denominator; thus $a$ is not a positive integer, which is a contradiction; thus there exist no solutions over the positive integers for $a$ and $b$ , and we are done.

Note: This problem is likely wrong.

~ eevee9406

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1989 OIM Problems/Problem 6

Problem

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