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Number Theory Two identities

Albert

Well-known member
Jan 25, 2013
1,225
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)



 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: help ! for these two identities

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

Albert

Well-known member
Jan 25, 2013
1,225

mente oscura

Well-known member
Nov 29, 2013
172
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.

[tex]x=\dfrac{m^2+mn+n^2}{4mn-1}=[/tex]

[tex]=\dfrac{m^2+mn+n^2}{4mn-1+4m^2-4m^2+4n^2-4n^2}=[/tex]

[tex]=\dfrac{1}{4-\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}}[/tex]

[tex]4>\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}\geq{3}[/tex]

[tex]4m^2+4n^2+1 \geq{3m^2+3mn+3n^2}[/tex]

[tex]m^2+n^2 \geq{3mn-1}[/tex]

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

[tex]x \le{1}[/tex]

Regards.
 

Albert

Well-known member
Jan 25, 2013
1,225
Hello.

[tex]x=\dfrac{m^2+mn+n^2}{4mn-1}=[/tex]

[tex]=\dfrac{m^2+mn+n^2}{4mn-1+4m^2-4m^2+4n^2-4n^2}=[/tex]

[tex]=\dfrac{1}{4-\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}}[/tex]

[tex]4>\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}\geq{3}[/tex]

[tex]4m^2+4n^2+1 \geq{3m^2+3mn+3n^2}[/tex]

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

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

[tex]x \le{1}[/tex]

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

mente oscura

Well-known member
Nov 29, 2013
172
[tex]4>\dfrac{4m^2+4n^2+1}{m^2+mn+n^2}\geq{3}[/tex]
[tex]4m^2+4n^2+1 \geq{3m^2+3mn+3n^2}[/tex]
[tex]m^2+n^2 \geq{3mn-1}---(*)[/tex]
from(*)
(**)should be :
[tex]x=\dfrac{m^2+mn+n^2}{4mn-1} \ge{\dfrac{3mn-1+mn}{4mn-1}}[/tex]
this implies :
[tex]x \ge{1}[/tex]
now you have to prove x=1
Hello.
Really, there is a mistake of sign, ultimately. I am sorry.:mad:

I continue in it.

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

[tex]4|(d^2+3) \ and \ 8\cancel{|}(d^2+3)[/tex]

Regards.
 

mente oscura

Well-known member
Nov 29, 2013
172
Hello.

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

[tex]x=\dfrac{m^2+mn+n^2}{4mn-1}[/tex]

[tex]m^2-n(4x-1)m+n^2+x=0[/tex]

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

Then:

[tex]pq=n^2+x \rightarrow{n^2=pq-x}[/tex]

[tex]p+q=n(4x-1)[/tex]

General demostration:

[tex](p+q)^2=n^2(4x-1)^2[/tex]

[tex](p+q)^2=(pq-x)(4x-1)^2[/tex]

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

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

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

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

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

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

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

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

Regards.
 

mente oscura

Well-known member
Nov 29, 2013
172
(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)

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

Demostratión:

[tex]Let \ d=2n-1, \ for \ n\in{N}[/tex]

[tex]d^2+3=(2n-1)^2+3=4n^2-4n+1+3=4(n^2-n+1)[/tex]

2º)

[tex]\dfrac{d^2+3}{b^2+bd-1}[/tex]

[tex]yb^2+ydb-y-d^2-3=0[/tex]

[tex]Let \ "p" \ and \ "q" \ roots \ with \ relation \ to \ "b"[/tex]

[tex]p+q=-yd[/tex]

[tex]pq=-y-d^2-3[/tex]

General demonstration:

[tex](p+q)^2=y^2d^2[/tex]

[tex](p+q)^2=y^2(-pq-y-3)=-y^2(pq+3)-y^3[/tex]

[tex]F(y)=y^3+(pq+3)y^2+(p+q)^2=0[/tex]

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

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

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

[tex]y=\dfrac{-(pq+7)\pm\sqrt{(pq+7)^2-4(4pq+28)}}{2}[/tex]


[tex]If \ \sqrt{(pq+7)^2-4(4pq+28)} \in{Z}[/tex]:

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:

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

Regards.