# 1984 IMO Problems/Problem 6

## Problem

Let $a,b,c,d$ be odd integers such that $0 and $ad=bc$. Prove that if $a+d=2^k$ and $b+c=2^m$ for some integers $k$ and $m$, then $a=1$.

## Solution 1

Let $f:[1,b]\rightarrow \mathbb{R},\ f(x)=x+\dfrac{bc}{x}$. As $f^\prime (x)=1-\dfrac{bc}{x^2}\le 0$, we infer that $f(x)\ge f(b)=b+c,\ \forall x\in [1,b]$; in particular, $a+d=f(a)\ge b+c\Leftrightarrow k\ge m$.

Now, $ad=bc\Leftrightarrow a(2^k-a)=b(2^m-b)\Leftrightarrow (b-a)(a+b)=2^m(b-2^{k-m}a)$, hence $2^m|(b-a)(a+b)$. It is easy to see that for $x,y\in \mathbb{Z}$, if $v_2(x\pm y)\ge 2$, then $v_2(x\mp y)=1$. If $v_2(b-a)\ge 2$, $v_2(a+b)=1$, so $v_2(b-a)\ge m-1\Rightarrow b>2^{m-1}$, which is in contradiction with the fact that $b<\dfrac{b+c}{2}=2^{m-1}$. Thereby, $v_2(a+b)\ge m-1,\ v_2(b-a)=1$.

Write $a+b=2^{m-1}\alpha$. If $\alpha \ge 2\Rightarrow2^m\le a+b, contradiction; so $a+b=2^{m-1}$ and $b-a=2\beta$, or equivalently $a=2^{m-2}-\beta,\ b=2^{m-2}+\beta$ ( $m>2$ otherwise $b+c=2\Leftrightarrow b=c=1$ or $b+c=4\Leftrightarrow c=3,b=1\Rightarrow a=0$ , contradiction )

Substituting back, $(b-a)(a+b)=2^m(b-2^{k-m}a)\Leftrightarrow 2^m\beta=2^m(2^{m-2}+\beta-2^{k-m}a)\Leftrightarrow 2^{k-m}a=2^{m-2}$

As $m>2$ and $a$ is odd, we get that $a=1$. Furthermore, $k=2m-2$, hence $d=2^{2m-2}-1$. Now $b(2^m-b)=2^{2m-2}-1\Leftrightarrow \left ( b-(2^{m-1}-1) \right ) \left ( b-(2^{m-1}+1)\right )=0$ which together with the fact that $b, we get that $b=2^{m-1}-1,\ c=2^{m-1}+1$, so the family of the solutions $(a,b,c,d)$ is described by the set $$M=\{ \left (1,2^{m-1}-1,2^{m-1}+1,2^{2m-2}-1 \right )|\ m\in \mathbb{N},m\ge 3 \}$$

This solution was posted and copyrighted by TheFunkyRabbit. The original thread for this problem can be found here: [1]

## Solution 2

Note that, if $k\leqslant m$, then using $a+d\leqslant b+c$, and $ad=bc$, we get $(d-a)^2\leqslant (c-b)^2$, which is clearly impossible. Thus, $k>m$ must hold.

Next, observe that, $b+c =2^m$ and $b implies $b<2^{m-1}$. Keeping this in mind, we now observe that, $bc=b(2^m-b)=ad=a(2^k-a)$, implies $b\cdot 2^m - a\cdot 2^k = (b-a)(b+a)$. Thus, $2^m \mid (b-a)(b+a)$. Now, note also that, if $d={\rm gcd}(b-a,b+a)$, then $d\mid 2b$, and thus, the largest power of $2$ dividing either $b-a$ or $b+a$ is exactly $2$. Clearly, $b-a<2^{m-1}$, and thus, $2^{m-1}\mid b+a$. Moreover, if $b+a\geqslant 2\cdot 2^{m-1}$, then we again have a contradiction, as $b,a<2^{m-1}$. Thus, $b+a=2^{m-1}$. This yields, $b-a = 2(b-a\cdot 2^{k-m})$, which brings us, $b=(2^{k-m+1}-1)a$. Now, using $b+a=2^{k-m+1}a = 2^{m-1}$, we immediately obtain $a=2^{2m-k-2}$. This immediately establishes (since $a$ is odd), that $a=1$.

This solution was posted and copyrighted by grupyorum. The original thread for this problem can be found here: [2]