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Difference between revisions of "1989 OIM Problems/Problem 6"

(Created page with "== Problem == Prove that there's and infinity of pairs of natural numbers that satisfy the equation: <math>2x^2-3x=3y^2</math> ~translated into English by Tomas Diaz. ~orders...")
 
(solution)
 
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== Solution ==
 
== Solution ==
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{{solution}}
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We claim that there do <b>not</b> exist any solutions to this equation over the positive integers.
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Assume for the sake of contradiction that there exists a solution over the positive integers. First, the equation implies that <math>2x^2=3(x+y^2)</math>; clearly then <math>3|x</math>, so let <math>x=3a</math>. Then the equation, after simplification, becomes <math>6a^2=3a+y^2</math>. Now <math>3|y</math> from this equation, so let <math>y=3b</math>; therefore, <math>2a^2=a+3b^2</math> for positive integers <math>a</math> and <math>b</math>.
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Solving the quadratic in <math>a</math> yields
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<cmath>a=\frac{1\pm\sqrt{1+24b^2}}{4}</cmath>
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Clearly, for <math>a</math> to be rational, let alone an integer, <math>\sqrt{1+24b^2}</math> must be an integer. Let <math>z=\sqrt{1+24b^2}</math>; then <math>z^2-24b^2=1</math>. This is the Pell Equation for <math>n=24</math>. Testing values yields the minimal solution <math>(z_1,b_1)=(5,1)</math>. (Also notice that <math>z^2=24b^2+1\ge24+1=25</math> since <math>b</math> is a positive integers, hence <math>z\ge5</math>.) Then all solutions <math>z_k</math> in <math>(z_k,b_k)</math> can be generated by the recurrence
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<cmath>z_k=z_1z_{k-1}+24b_1b_{k-1}</cmath>
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Taking the equation <math>\pmod{4}</math>:
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<cmath>z_k\equiv z_1z_{k-1}\pmod{4}</cmath>
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We showed earlier that <math>z_1=5</math>; hence,
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<cmath>z_k\equiv 5z_{k-1}\equiv z_{k-1}\pmod{4}</cmath>
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for all integers <math>k\ge2</math>. Since <math>z_1\equiv1\pmod{4}</math>, we can use the recurrence to conclude that
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<cmath>z_k\equiv1\pmod{4}</cmath>
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for all such <math>k</math>. Therefore, all integer values of the square root must have this be true. Thus,
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<cmath>a=\frac{1\pm z_k}{4}</cmath>
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But notice that the numerator must be positive, and since <math>z_k</math> is an integer, the <math>\pm</math> must be a <math>+</math>, so
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<cmath>a=\frac{1+z_k}{4}</cmath>
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However, <math>1+z_k\equiv2\pmod{4}</math>, so the numerator is not divisible by the denominator; thus <math>a</math> is not a positive integer, which is a contradiction; thus there exist no solutions over the positive integers for <math>a</math> and <math>b</math>, and we are done.
 +
 
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<i>Note: This problem is likely wrong.</i>
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe4.htm
 
https://www.oma.org.ar/enunciados/ibe4.htm

Latest revision as of 18:43, 14 April 2025

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

See also

https://www.oma.org.ar/enunciados/ibe4.htm

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