Let $a, b, c, d$ be odd integers where $0 \lt a \lt b \lt c \lt d$. It is known that $ad=bc$ while $a+d$ and $b+c$ are powers of $2$, i.e., there exist positive integers $x,y$ such that $a+d=2^x$ and $b+c=2^y$. Show that $a=1$
I have tried replacing d=$\frac{bc}{a}$ etc., but it does not work.
Can someone give me a hint?
On
There are infinite results that can fix into your equation, infant is a Diophantine equation
$(a,b,c,d) € odd integers$
Where $0<a<b<c<d$ and $a*d = b*c$ and $a+d = 2^x$ and $b+c = 2^y$, where $(x,y) € Z$
The set of odd integers include
$(1,3,5,7,9,11,13,15,17,19,21,23,25,27,......)$
If we imagine that $a=1$, just to create the smallest solutions to the problem
Since $a+d = 2^x$ They'll be grouped in this way
$[ (1,3),(1,7),(1,15),(1,31),(1,63),..... ]$
Since $a*d = b*c$, their product must be equal to the product of two other odd numbers, therefore will exclude the prime therein
Then the first set of solutions for $a$ and $d$ are $:$
$[ (1,15),(1,63),(1,255),(1,1023),....... ]$
By the splitting of this numbers, we create the values of $b$ and $c$, because $15=3×5$, $63=7×9$, $255=15×17$, $1023=33×31$
Then the simplest-case solutions are
$(1,3,5,15)$
$(1,7,9,63)$
$(1,15,17,255)$
$(1,31,33,1023)$ e.t.c
Now let's imagine $a$ was not $1$, say $a=3$ (the next odd number)
So the list of the sets that combines to make $a+d = 2^x$ are
$[ (3,5),(3,13),(3,29),(3,61),(3,125),...... ]$
Since $a*d = b*c$, their product must equal to the product of two other smaller odd numbers, remember that $0<a<b<c<d$, so we'll exclude the primes Therefore the solutions for $a$ and $d$ here are
$[ (3,125),(3,253),(3,2045),........... ]$
By splitting up this numbers, when create the values for $b$ and $c$
Because $125=5*25$, $253=11×23$, $2045=5×409$,........
A problem then occurs it wouldn't satisfy that $b+c = 2^y$
This results are wrong $(3,15,25,125)$ $(3,5,75,125)$ $(3,23,33,253)$ $(3,11,69,253)$ $(3,5,1227,2045)$ etc
So maybe the only value of $a$ that works is when it's equal to $1$
Proofing it takes a different kind of maths
I expect there is a more elegant solution to this, but at least this works (barring arithmetic error which, of course, is always possible):
First remark that we must have $a+d>b+c$:
Pf: Indeed $$c(b-a)=bc-ac=ad-ac=a(d-c)<c(d-c)\implies c(b-a)<c(d-c)\implies $$$$\implies (b-a)<(d-c)\implies (b+c)<(a+d)$$ as desired.
It follows that $x>y$.
Now write $d=2^x-a$ and $c=2^y-b$ to deduce that $$ad=bc\implies 2^xa-a^2=2^yb-b^2\implies 2^y(b-2^{x-y}a)=(b+a)(b-a)$$
Since $a,b$ are both odd we must have that $2^{y-1}$ divides one of $b+a$ and $b-a$ (since $\gcd (b+a,b-a)$ divides $2b$ and $b$ is odd).
Case I: $2^{y-1}\,|\,(b-a)$ . This is absurd since it would imply that $b>2^{y-1}$ which would imply that $c>2^{y-1}$ and then $b+c>2^{y}$ which is false.
Case II: $2^{y-1}\,|\,(b+a)$ So $(b+a)=2^{y-1}n$. If $n≥2$ then we'd have $(b+a)≥(c+a)$ which is false so we must have $n=1$. Hence $$b+a=2^{y-1}$$
Finally we can put all this together:
$$a=2^{y-1}-b\quad \&\quad c=2^y-b\implies c=2^{y-1}+a$$ Hence $$bc=ad\implies (2^{y-1}-a)(2^{y-1}+a)=a(2^x-a)\implies 2^{2y-2}-a^2=2^xa-a^2\implies $$ $$\implies 2^{2y-2}=2^xa$$ which, finally, implies that $a=1$.