2015 OIM Problems/Problem 5

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

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2015 OIM Problems/Problem 5

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