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Difference between revisions of "2015 OIM Problems/Problem 5"

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== Solution ==
 
== Solution ==
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{{solution}}
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First we deal with the zero cases. Assume that <math>a=0</math>; then the right-hand side (RHS) is zero, so we must have <math>b^2-7b=0</math>; therefore our solutions are <math>(0,0)</math> and <math>(0,7)</math>. These both clearly work. Next assume that <math>b=0</math>; then the RHS is also zero, so <math>7a=0</math> and thus <math>a=0</math>, which we already considered.
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Next, we consider nonzero <math>a,b</math>. We can rewrite the RHS as <math>a^2\cdot ab</math>; then <math>ab</math> must then be a perfect square, so let <math>ab=k^2</math> for some integer <math>k</math>. Then the equation becomes:
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<cmath>(b^2+7(a-b))^2=a^2k^2</cmath>
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<cmath>b^2+7(a-b)=ak</cmath>
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(Both positive and negative cases are already considered since <math>k</math> can be either positive or negative.)
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 +
Our equation can be rearranged to become:
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<cmath>b(b-7)=a(k-7)</cmath>
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Multiplying both sides by <math>b</math>:
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<cmath>b^2(b-7)=ab(k-7)=k^2(k-7)</cmath>
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This can be rearranged to produce:
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<cmath>b^3-k^3=7(b^2-k^2)</cmath>
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Assume that <math>b=k</math>. Then <math>b^2=ab</math>, and since <math>b</math> is nonzero, we must have <math>a=b</math>. All such solutions work in the equation.
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Next, assume that <math>b\ne k</math>. Then we can divide <math>(b-k)</math> from both sides:
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<cmath>(b-k)(b^2+bk+k^2)=7(b+k)(b-k)</cmath>
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<cmath>b^2+bk+k^2=7b+7k</cmath>
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<cmath>b^2+(k-7)b+(k^2-7k)=0</cmath>
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Using the [[Quadratic Formula]]:
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<cmath>b=\frac{-k+7\pm\sqrt{(k-7)^2-4(k^2-7k)}}{2}=\frac{-k+7\pm\sqrt{-3k^2+14k+49}}{2}=\frac{-k+7\pm\sqrt{(3k+7)(7-k)}}{2}</cmath>
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Consider the discriminant. We must have <math>(3k+7)(7-k)\ge0</math>, so we have to consider integers <math>k\in[-2,7]</math>. From casework we derive that <math>k=-2,3,6,7</math> all cause the discriminant to be a perfect square, respectively <math>9,64,25,0</math>. Solving the equation yields the pairs
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<cmath>k=-2,b=3,6</cmath>
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<cmath>k=3,b=-2,6</cmath>
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<cmath>k=6,b=-2,3</cmath>
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<cmath>k=7,b=0</cmath>
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<math>b\ne0</math>, so we can eliminate the final pair. Otherwise, since <math>k^2=ab</math>, we must have that <math>b|k^2</math>. This leaves
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<cmath>k=6,b=-2,3</cmath>
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as the only possibility. From here we find that <math>a=-18,12</math> respectively, so our pairs are <math>(-18,-2)</math> and <math>(12,3)</math>, which both work.
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Notice that <math>(0,0)</math> is a member of the set of points <math>a=b</math>, so to sum up, the solutions to the equation are <math>\boxed{(0,7),(-18,-2),(12,3),(t,t)}</math> for integers <math>t</math>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
[[OIM Problems and Solutions]]
 
[[OIM Problems and Solutions]]

Latest revision as of 18:10, 8 April 2025

Problem

Find all pairs $(a, b)$ of integers that verify:

\[(b^2+7(a-b))^2=a^3b\]

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

Solution

First we deal with the zero cases. Assume that $a=0$; then the right-hand side (RHS) is zero, so we must have $b^2-7b=0$; therefore our solutions are $(0,0)$ and $(0,7)$. These both clearly work. Next assume that $b=0$; then the RHS is also zero, so $7a=0$ and thus $a=0$, which we already considered.

Next, we consider nonzero $a,b$. We can rewrite the RHS as $a^2\cdot ab$; then $ab$ must then be a perfect square, so let $ab=k^2$ for some integer $k$. Then the equation becomes: \[(b^2+7(a-b))^2=a^2k^2\] \[b^2+7(a-b)=ak\] (Both positive and negative cases are already considered since $k$ can be either positive or negative.)

Our equation can be rearranged to become: \[b(b-7)=a(k-7)\] Multiplying both sides by $b$: \[b^2(b-7)=ab(k-7)=k^2(k-7)\] This can be rearranged to produce: \[b^3-k^3=7(b^2-k^2)\] Assume that $b=k$. Then $b^2=ab$, and since $b$ is nonzero, we must have $a=b$. All such solutions work in the equation.

Next, assume that $b\ne k$. Then we can divide $(b-k)$ from both sides: \[(b-k)(b^2+bk+k^2)=7(b+k)(b-k)\] \[b^2+bk+k^2=7b+7k\] \[b^2+(k-7)b+(k^2-7k)=0\] Using the Quadratic Formula: \[b=\frac{-k+7\pm\sqrt{(k-7)^2-4(k^2-7k)}}{2}=\frac{-k+7\pm\sqrt{-3k^2+14k+49}}{2}=\frac{-k+7\pm\sqrt{(3k+7)(7-k)}}{2}\] Consider the discriminant. We must have $(3k+7)(7-k)\ge0$, so we have to consider integers $k\in[-2,7]$. From casework we derive that $k=-2,3,6,7$ all cause the discriminant to be a perfect square, respectively $9,64,25,0$. Solving the equation yields the pairs \[k=-2,b=3,6\] \[k=3,b=-2,6\] \[k=6,b=-2,3\] \[k=7,b=0\] $b\ne0$, so we can eliminate the final pair. Otherwise, since $k^2=ab$, we must have that $b|k^2$. This leaves \[k=6,b=-2,3\] as the only possibility. From here we find that $a=-18,12$ respectively, so our pairs are $(-18,-2)$ and $(12,3)$, which both work.

Notice that $(0,0)$ is a member of the set of points $a=b$, so to sum up, the solutions to the equation are $\boxed{(0,7),(-18,-2),(12,3),(t,t)}$ for integers $t$.

~ eevee9406

See also

OIM Problems and Solutions

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