$a+d=2^k,b+c=2^m$,proving $a=1$ $a<b<c<d$ 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 ?
 A: It will simplify the exposition a bit if we allow reordering of $b,c$, that is, $a<b<c<d$ or $a<c<b<d$. 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 $a<\{b\ne c\}<d$ 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
$$
r+q<3\cdot 2^{m-1}, ~\mathrm{so}~2^{m-1}\mid r+q \Rightarrow r+q=2^{m-1}~\mathrm{or}~r+q=2^m \\
r+q=2^m \Rightarrow p=s=1 \Rightarrow a+b=c+d~\mathrm{which~cannot~be} \\
\mathrm{so}~~ r+q=2^{m-1}\\
-2^{m-1}<s-p<2^m, ~\mathrm{so}~2^{m-1}\mid s-p \Rightarrow s-p=2^{m-1}~\mathrm{or}~s-p=0 \\
s-p=0 \Rightarrow d=b ~\mathrm{which~cannot~be} \\
\mathrm{so}~~ s-p=2^{m-1}\\
$$
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 $r<s$ 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
$$
a=1, b=2^{m}-1, c=2^{m}+1, d=2^{2m}-1
$$
