2004 OIM Problems/Problem 4

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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{align*} a&=\frac{602\pm\sqrt{602^2-4(101^2-4b)}}{2}\\ &=\frac{602\pm2\sqrt{301^2-(101^2-4b)}}{2}\\ &=\frac{602\pm2\sqrt{4b+301^2-101^2}}{2}\\ &=\frac{602\pm2\sqrt{4b+402\cdot200}}{2}\\ &=\frac{602\pm4\sqrt{b+402\cdot50}}{2}\\ &=\frac{602\pm4\sqrt{b+20100}}{2}\\ \end{align*} 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{align*} a&=\frac{602\pm4\sqrt{b+20100}}{2}\\ &=\frac{602\pm4\cdot142}{2}\\ &=\frac{602\pm568}{2}\\ &=\frac{602-568}{2}\\ &=\frac{34}{2}\\ &=17\\ \end{align*} 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

See also

OIM Problems and Solutions