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$a are odd natural numbers. $ad=bc$

$a+d=2^{k},b+c=2^{m}$ how to prove that $a=1$

I heard that we can prove it by "Four Number Theorem" ,is that right?

And is there a different way to prove it ?

1 Answers 1

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It will simplify the exposition a bit if we allow reordering of $b,c$, that is, $a or $a. Clearly they can be interchanged without changing the problem.

The Four Number Theorem tells us that there are positive integers $p,q,r,s$ with $a=pq, d=rs, b=pr, c=qs$. Since $a,b,c,d$ are odd so must be $p,q,r,s$. $ \frac{a+d+b+c}{4} = \left(\frac{s+p}{2}\right)\left(\frac{r+q}{2}\right) = 2^{m-2}(2^{k-m}+1) \\ \frac{a+d-b-c}{4} = \left(\frac{s-p}{2}\right)\left(\frac{r-q}{2}\right) = 2^{m-2}(2^{k-m}-1) $

From $a<\{b\ne c\} all odd we have $b+c\ge 8$, so $2^{m-2}$ is even and if one of the factors in the middle representations above is odd then $2^{m-2}$ must divide the other factor.

$a+d\equiv 0 \pmod{4}\Rightarrow ad=pqrs\equiv -1\pmod{4}$ so one or three of $p,q,r,s$ are congruent to $-1\pmod 4$. The possibilities are $ \begin{array}{lrc} \mathbf{either} & p\equiv s\pmod{4} \Rightarrow &\frac{p+s}{2}~ \mathrm{is~odd} \Rightarrow 2^{m-2}\mid \frac{r+q}{2} \\ & \mathrm{and}&\frac{r-q}{2}~\mathrm{is~odd}\Rightarrow 2^{m-2}\mid \frac{s-p}{2} \\ \mathbf{or} & r\equiv q\pmod{4} \Rightarrow & \frac{r+q}{2}~ \mathrm{is~odd} \Rightarrow 2^{m-2}\mid \frac{s+p}{2} \\ & \mathrm{and}&\frac{s-p}{2}~\mathrm{is~odd}\Rightarrow 2^{m-2}\mid \frac{r-q}{2} \end{array} $

We have $ p,q\le a < \min(b,c) < \frac{b+c}{2} = 2^{m-1},~~ r\le b<2^m,~~ s\le c<2^m $ so each of the cases above proceeds as follows $ r+q<3\cdot 2^{m-1}, ~\mathrm{so}~2^{m-1}\mid r+q \Rightarrow r+q=2^{m-1}~\mathrm{or}~r+q=2^m \\ r+q=2^m \Rightarrow p=s=1 \Rightarrow a+b=c+d~\mathrm{which~cannot~be} \\ \mathrm{so}~~ r+q=2^{m-1}\\ -2^{m-1} So either $r+q=s-p=2^{m-1}$ or $s+p=r-q=2^{m-1}$.

We now repeat the entire argument with the alternative representations $ a+d+b+c=(s+q)(r+p)\\ a+d-b-c=(s-q)(r-p) $ to get $r+p=s-q=2^{m-1}$ or $s+q=r-p=2^{m-1}$. Matching the cases on $r>s$ or $r either way we get $s+r=2^m=b+c\Rightarrow p=q=1=a$.

This also illustrates that the possible solutions are of the form $ a=1, b=2^{m}-1, c=2^{m}+1, d=2^{2m}-1 $