MHB Proving Equality for $x$ and $y$

[Albert1]

help ! for these two identies

(1)$m,n \in N$
$x=\dfrac {m^2+mn+n^2}{4mn-1}$
if $x\in N$
prove: x=1
(ex:m=2, n=1)
(2)$b,d \in N$
b (being even numbers)
d (being odd numbers)
$y=\dfrac {d^2+3}{b^2+bd-1}$
if $y\in N$
prove: y=4
(ex:b=2, d=9)
​

 

[Prove It]

Re: help ! for these two identities

Are you really trying to prove x = 1 and y = 4?

 

[Albert1]

Re: help ! for these two identities

Are you really trying to prove x = 1 and y = 4?

yes ,please

 

[mente oscura]

help ! for these two identies

(1)$m,n \in N$
$x=\dfrac {m^2+mn+n^2}{4mn-1}$
if $x\in N$
prove: x=1
(ex:m=2, n=1)
​

Hello.

x=\dfrac{m^2+mn+n^2}{4mn-1}=

=\dfrac{m^2+mn+n^2}{4mn-1+4m^2-4m^2+4n^2-4n^2}=

=\dfrac{1}{4-\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}}

4>\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}\geq{3}

4m^2+4n^2+1 \geq{3m^2+3mn+3n^2}

m^2+n^2 \geq{3mn-1}

x=\dfrac{m^2+mn+n^2}{4mn-1} \le{\dfrac{3mn-1+mn}{4mn-1}}

x \le{1}

Regards.

 

[Albert1]

Hello.

x=\dfrac{m^2+mn+n^2}{4mn-1}=

=\dfrac{m^2+mn+n^2}{4mn-1+4m^2-4m^2+4n^2-4n^2}=

=\dfrac{1}{4-\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}}

4>\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}\geq{3}

4m^2+4n^2+1 \geq{3m^2+3mn+3n^2}

m^2+n^2 \geq{3mn-1}---(*)

x=\dfrac{m^2+mn+n^2}{4mn-1} \le{\dfrac{3mn-1+mn}{4mn-1}}---(**)

x \le{1}

Regards.

4>\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}\geq{3}
4m^2+4n^2+1 \geq{3m^2+3mn+3n^2}
m^2+n^2 \geq{3mn-1}---(*)
from(*)
(**)should be :
x=\dfrac{m^2+mn+n^2}{4mn-1} \ge{\dfrac{3mn-1+mn}{4mn-1}}
this implies :
x \ge{1}
now you have to prove x=1

 

[mente oscura]

4>\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}\geq{3}
4m^2+4n^2+1 \geq{3m^2+3mn+3n^2}
m^2+n^2 \geq{3mn-1}---(*)
from(*)
(**)should be :
x=\dfrac{m^2+mn+n^2}{4mn-1} \ge{\dfrac{3mn-1+mn}{4mn-1}}
this implies :
x \ge{1}
now you have to prove x=1

Hello.
Really, there is a mistake of sign, ultimately. I am sorry.

I continue in it.

In all that, to the second topic, I do not manage to advance any more that:

4|(d^2+3) \ and \ 8\cancel{|}(d^2+3)

Regards.

 

[mente oscura]

Hello.

Good, I am going to try it again. I wait not to have been wrong in the calculations.

x=\dfrac{m^2+mn+n^2}{4mn-1}

m^2-n(4x-1)m+n^2+x=0

Let \ p, \ q \in{R}/ roots \ of \ m

Then:

pq=n^2+x \rightarrow{n^2=pq-x}

p+q=n(4x-1)

General demostration:

(p+q)^2=n^2(4x-1)^2

(p+q)^2=(pq-x)(4x-1)^2

F(x)=16x^3-(16pq+8)x^2+(8pq+1)x+(p+q)^2-pq=0

Since we know, that "1" is a root:

\dfrac{F(x)}{x-1}=16x^2-(16pq-8)x-(8pq-9)+Rest

If "1" is a root, then Rest=0

x=\dfrac{16pq-8\pm\sqrt{(16pq-8)^2+64(8pq-9)}}{32}

x=\dfrac{2pq-1\pm\sqrt{4p^2q^2+4pq-8}}{4}

Therefore, "x" does not have any more entire roots.

If it is correct, I will go for the second question. :p

Regards.

 

[mente oscura]

(2)$b,d \in N$
b (being even numbers)
d (being odd numbers)
$y=\dfrac {d^2+3}{b^2+bd-1}$
if $y\in N$
prove: y=4
(ex:b=2, d=9)

Hello.

I am going to try to solve the second question.

1º)

If \ d\in{N} \ /d=odd \rightarrow{4|(d^2+3) \ and \ 8\cancel{|}(d^2+3)}

Demostratión:

Let \ d=2n-1, \ for \ n\in{N}

d^2+3=(2n-1)^2+3=4n^2-4n+1+3=4(n^2-n+1)

2º)

\dfrac{d^2+3}{b^2+bd-1}

yb^2+ydb-y-d^2-3=0

Let \ "p" \ and \ "q" \ roots \ with \ relation \ to \ "b"

p+q=-yd

pq=-y-d^2-3

General demonstration:

(p+q)^2=y^2d^2

(p+q)^2=y^2(-pq-y-3)=-y^2(pq+3)-y^3

F(y)=y^3+(pq+3)y^2+(p+q)^2=0

Since we know, that "4" is a root:

\dfrac{F(y)}{y-4}=y^2+(pq+7)y+(4pq+28)+Rest

If "4" is a root with relation to "y", then Rest=0

y=\dfrac{-(pq+7)\pm\sqrt{(pq+7)^2-4(4pq+28)}}{2}If \ \sqrt{(pq+7)^2-4(4pq+28)} \in{Z}:

A contradiction would happen, as for the paragraph 1 º, since one of the numerical roots might be divisible only for "4", but other one would be wholesale or minor of "4".

Conclusion:

\sqrt{(pq+7)^2-4(4pq+28)} \cancel{\in}{Z}:)

Regards.