IMO 2019 Shortlisted Problems

Assume, for contradiction, that $$a^2+\lceil \frac{4a^2}{b}\rceil =(a+k{)}^2,\text{or}\lceil \frac{(2a{)}^2}{b}\rceil =(2a+k)k.$$ Clearly, $k⩾1$. In other words, the equation $$\lceil \frac{c^2}{b}\rceil =(c+k)k\text{\hspace{1em}(1)}$$ has a positive integer solution $(c,k)$, with an even value of $c$.

Choose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From $$\frac{c^2}{b}>\lceil \frac{c^2}{b}\rceil -1=ck+k^2-1⩾ck$$ and $$\frac{(c-k)(c+k)}{b}<\frac{c^2}{b}⩽\lceil \frac{c^2}{b}\rceil =(c+k)k$$ it can be seen that $c>bk>c-k$, so $$c=kb+r\text{with some}0<r<k.$$ By substituting this in (1) we get $$\lceil \frac{c^2}{b}\rceil =\lceil \frac{(bk+r{)}^2}{b}\rceil =k^{2}b+2kr+\lceil \frac{r^2}{b}\rceil$$ and $$(c+k)k=(kb+r+k)k=k^{2}b+2kr+k(k-r),$$ so $$\lceil \frac{r^2}{b}\rceil =k(k-r)\text{\hspace{1em}(2)}$$ Notice that relation (2) provides another positive integer solution of (1), namely $c^{\prime }=r$ and $k^{\prime }=k-r$, with $c^{\prime }>0$ and $0<k^{\prime }<k$. That contradicts the minimality of $k$, and hence finishes the solution.

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Alternative solution.

Suppose that $$a^2+\lceil \frac{4a^2}{b}\rceil =c^2$$ with some positive integer $c>a$, so $$\begin{array}{rl}& c^2-1<a^2+\frac{4a^2}{b}⩽c^2\\ & 0⩽c^{2}b-a^2(b+4)<b\end{array}\text{\hspace{1em}(3)}$$ Let $d=c^{2}b-a^2(b+4)$, $x=c+a$ and $y=c-a$; then we have $c=\frac{x+y}{2}$ and $a=\frac{x-y}{2}$, and (3) can be re-written as follows: $$\begin{array}{rl}{\left(\frac{x+y}{2}\right)}^{2}b-{\left(\frac{x-y}{2}\right)}^2(b+4)& =d,\\ x^2-(b+2)xy+y^2+d=0;0& ⩽d<b.\end{array}\text{\hspace{1em}(4)}$$ So, by the indirect assumption, the equation (4) has some positive integer solution $(x,y)$.

Fix $b$ and $d$, and take a pair $(x,y)$ of positive integers, satisfying (4), such that $x+y$ is minimal. By the symmetry in (4) we may assume that $x⩾y⩾1$.

Now we perform a usual "Vieta jump". Consider (4) as a quadratic equation in variable $x$, and let $z$ be its second root. By the Vieta formulas, $$x+z=(b+2)y,\text{and}zx=y^2+d$$ so $$z=(b+2)y-x=\frac{y^2+d}{x}$$ The first formula shows that $z$ is an integer, and by the second formula $z$ is positive. Hence $(z,y)$ is another positive integer solution of (4). From $$\begin{array}{rl}(x-1)(z-1)& =xz-(x+z)+1=(y^2+d)-(b+2)y+1\\ & <(y^2+b)-(b+2)y+1=(y-1{)}^2-b(y-1)⩽(y-1{)}^2⩽(x-1{)}^2\end{array}$$ we can see that $z<x$ and therefore $z+y<x+y$. But this contradicts the minimality of $x+y$ among the positive integer solutions of (4).