MHB Proving $m+n=xy$ Using Positive Real Numbers

[anemone]

Let $x,\,y,\,m,\,n$ be positive real numbers such that $m^2-m+1=x^2$, $n^2+n+1=y^2$ and $(2m-1)(2n+1)=2xy+3$.

Prove that $m+n=xy$.

 

[Albert1]

Let $x,\,y,\,m,\,n$ be positive real numbers such that $m^2-m+1=x^2---(1)$, $n^2+n+1=y^2---(2)$

and $(2m-1)(2n+1)=2xy+3---(3)$.

Prove that $m+n=xy---(4)$.

Spoiler

let :$p=2m-1,q=2n+1$
if(4) is true then :
$pq-3=2xy=p+q=2(m+n)$
or $\Leftrightarrow (p-1)(q-1)=4=4n(m-1)$
or :$\Leftrightarrow n(m-1)=1---(5)$
put (5) to (1) and (2) we get :
$x^2=m^2-m+1$
$y^2=\dfrac {m^2-m+1}{(m-1)^2}$
$\therefore xy=\dfrac {m^2-m+1}{m-1}=m+\dfrac {1}{m-1}=m+n$
and the proof is done

 

[kaliprasad]

Spoiler

let :$p=2m-1,q=2n+1$
if(4) is true then :
$pq-3=2xy=p+q=2(m+n)$
or $\Leftrightarrow (p-1)(q-1)=4=4n(m-1)$
or :$\Leftrightarrow n(m-1)=1---(5)$
put (5) to (1) and (2) we get :
$x^2=m^2-m+1$
$y^2=\dfrac {m^2-m+1}{(m-1)^2}$
$\therefore xy=\dfrac {m^2-m+1}{m-1}=m+\dfrac {1}{m-1}=m+n$
and the proof is done

Spoiler

we are finding (5) from (4) and then we are proving (4). So solution seems to be incorrect. Or am I missing something.

 

[Albert1]

Spoiler

let :$p=2m-1,q=2n+1$
if(4) is true then :
$pq-3=2xy---(3)$
or $\Leftrightarrow (p-1)(q-1)=4=4n(m-1)$
or :$\Leftrightarrow n(m-1)=1---(5)$
put (5) to (1) and (2) we get :
$x^2=m^2-m+1$
$y^2=\dfrac {m^2-m+1}{(m-1)^2}$
$\therefore xy=\dfrac {m^2-m+1}{m-1}=m+\dfrac {1}{m-1}=m+n$
and the proof is done

Spoiler

let :$p=2m-1,q=2n+1, and \,\, n=\dfrac {1}{m-1}---(5) , (why?)$
put (5) to (3) we get :
$pq-3=2xy=2(m+n)=p+q$
$\therefore xy=m+n$
explanation of why :
only for $pq-3=p+q $ then $xy=m+n$ will satisfy
we get :$(p-1)(q-1)=4$ then $xy=m+n$
or $n(m-1)=1$ then $xy=m+n$

 

[anemone]

Hi Albert,

I'm unfamiliar with this type of backward proof, Albert...I don't know what to say as I don't find your proof convincing, sorry...:( but if I really have to prove it that way, perhaps I would first assume the equation (4) holds and then use the equation (3) to prove there exist both the equations (1) and (2) as stated by the original question to complete my proof...(all the numbering of the equations are based on your post #2)

This is an Olympiad math problem though...

Hint:

Spoiler

Find $(m+n)^2$ and $(xy)^2$ separately and then make appropriate comparison.

 

[kaliprasad]

Hi Albert,

I'm unfamiliar with this type of backward proof, Albert...I don't know what to say as I don't find your proof convincing, sorry...:( but if I really have to prove it that way, perhaps I would first assume the equation (4) holds and then use the equation (3) to prove there exist both the equations (1) and (2) as stated by the original question to complete my proof...(all the numbering of the equations are based on your post #2)

This is an Olympiad math problem though...

Hint:

Spoiler

Find $(m+n)^2$ and $(xy)^2$ separately and then make appropriate comparison.

I do not find the proof of Albert to be convincing and it is definitely with flaw

to prove A you assume B such that A true. He has taken a suitable value of n assuming the final result and based on that proved the final result. thus the final result is always true.

 

[Opalg]

Let $x,\,y,\,m,\,n$ be positive real numbers such that $m^2-m+1=x^2$, $n^2+n+1=y^2$ and $(2m-1)(2n+1)=2xy+3$.

Prove that $m+n=xy$.

[sp]Given $m^2-m+1=x^2$, $n^2+n+1=y^2$ and $(2m-1)(2n+1)=2xy+3$, it follows that $$2x^2 = 2m^2-2m+2,$$ $$2y^2 = 2n^2+2n+2,$$ $$2xy = 4mn + 2m - 2n - 4.$$ Add those three equations to get $2(x^2 + y^2 + xy) = 2m^2 + 4mn + 2n^2 = 2(m+n)^2$, so that $$ x^2 + y^2 + xy = (m+n)^2. \qquad(1)$$
From the given equation $m^2-m+1=x^2$, it follows that $4x^2 - 3 = 4m^2-4m+1 = (2m-1)^2$. Similarly, from $n^2+n+1=y^2$ it follows that $4y^2-3 = (2n+1)^2$. Multiply those equations together and use the given equation $(2m-1)(2n+1)=2xy+3$ to deduce that $$(4x^2-3)(4y^2-3) = (2xy+3)^2,$$ $$16x^2y^2 - 12x^2 - 12y^2 + 9 = 4x^2y^2 + 12xy + 9,$$ $$12 x^2y^2 - 12x^2 - 12y^2 = 12xy$$ and therefore $$x^2y^2 = x^2 + y^2 + xy.\qquad (2)$$
Now combine $(1)$ and $(2)$ to get $(m+n)^2 = x^2y^2$. Finally, we are told that all the numbers are positive, so we can take the positive square root of both sides to get $m+n=xy.$[/sp]

 

[anemone]

Thanks, Opalg, for participating!(Smile)