2004 OIM Problems/Problem 4

Problem

Find all pairs $(a, b)$ , where $a$ and $b$ are positive integers of two digits each, such that $100a + b$ and $201a + b$ are four-digit perfect squares.

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

Solution

Let $100a+b=x^2$ and $201a+b=y^2$ ; then, by subtracting, $y^2-x^2=101a$ , so $(y+x)(y-x)=101a$ . Notice that since $x^2$ and $y^2$ are four-digit, $x^2,y^2<10000$ , so $x,y<100$ . Thus $x+y<200$ .

Clearly $y>x$ , so both $y+x$ and $y-x$ are positive; however, as $y-x$ cannot equal $101$ due to $x,y<100$ , we must have $101|(y+x)$ . Since $x+y\ne0$ and $x+y<200<202$ , it is necessary that $x+y=101$ . Then $y-x=a$ . Solving for $x$ and $y$ results in $x=\frac{101-a}{2},y=\frac{101+a}{2}$ Substituting back in: $100a+b=\left(\frac{101-a}{2}\right)^2$ $201a+b=\left(\frac{101+a}{2}\right)^2$ From the second equation, rearranging yields $a^2-602a+101^2-4b=0$ Using the Quadratic Formula: $\begin{aligned} a & = \frac{602 \pm \sqrt{602^{2} - 4 (101^{2} - 4 b)}}{2} \\ = \frac{602 \pm 2 \sqrt{301^{2} - (101^{2} - 4 b)}}{2} \\ = \frac{602 \pm 2 \sqrt{4 b + 301^{2} - 101^{2}}}{2} \\ = \frac{602 \pm 2 \sqrt{4 b + 402 \cdot 200}}{2} \\ = \frac{602 \pm 4 \sqrt{b + 402 \cdot 50}}{2} \\ = \frac{602 \pm 4 \sqrt{b + 20100}}{2} \end{aligned}$ Clearly $b+20100$ must be a perfect square, and the only perfect square that allows $b$ to have two digits is $142^2=20164$ . Thus $b=64$ , and: $\begin{aligned} a & = \frac{602 \pm 4 \sqrt{b + 20100}}{2} \\ = \frac{602 \pm 4 \cdot 142}{2} \\ = \frac{602 \pm 568}{2} \\ = \frac{602 - 568}{2} \\ = \frac{34}{2} \\ = 17 \end{aligned}$ Thus the only solution is $(a,b)=\boxed{(17,64)}$ , which works (testing yields the two squares $42^2=1764$ and $59^2=3481$ ).

~ eevee9406

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2004 OIM Problems/Problem 4

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