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.