Square Numbers Satisfying $3x^2+x=4y^2+y$

  • Context: MHB 
  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Numbers Square
Click For Summary

Discussion Overview

The discussion revolves around the equation $3x^2+x=4y^2+y$ and seeks to demonstrate that for positive integer solutions $(x, y)$, certain expressions involving $x$ and $y$ yield perfect squares. The scope includes mathematical reasoning and exploration of integer solutions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Some participants assert that for positive integers $x$ and $y$ satisfying the equation, the expressions $x-y$, $3x+3y+1$, and $4x+4y+1$ must all be perfect squares.
  • One participant mentions that there is at least one known solution, specifically $(x, y) = (30, 26)$, which supports the exploration of the claims.
  • Another participant reiterates the need to show that $x-y$, $3x+3y+1$, and $4x+4y+1$ are squares, suggesting a focus on proving these relationships.

Areas of Agreement / Disagreement

Participants generally agree on the goal of demonstrating that certain expressions are squares for the given equation, but the discussion does not resolve the methods or approaches to achieve this.

Contextual Notes

The discussion does not clarify the assumptions or methods that participants might use to prove the claims, and it remains open regarding the completeness of the arguments presented.

Who May Find This Useful

Readers interested in number theory, particularly those exploring properties of integer solutions to polynomial equations, may find this discussion relevant.

kaliprasad
Gold Member
MHB
Messages
1,333
Reaction score
0
show that for x,y positive integers satisfying $3x^2+x= 4y^2+y$ each of x-y , 3x+3y+ 1 and 4x + 4y + 1 are squares. ( above equation has atleast one solution x= 30 and y = 26)
 
Mathematics news on Phys.org
kaliprasad said:
show that for x,y positive integers satisfying $3x^2+x= 4y^2+y--(1)$
each of x-y =a^2 (2),
3x+3y+ 1=b^2 (3)and
4x + 4y + 1 (4)
are squares. ( above equation has atleast one solution x= 30 and y = 26)
from (2):$x-y=a^2>0---(*)$
from (3):$x+y=\dfrac{b^2-1}{3}=\dfrac{(b+1)(b-1)}{3}$
$\therefore (b+1)\,\, mod \,\ 3=0\,\, or \,\, (b-1)\,\, mod \,\, 3=0$
if $x+y \,\, even \,\, then \,\, b \,\, must \,\, be\,\ odd$
take $b=13$,we have $x+y=56$
if we let $x-y=a^2=4,\,\, and \,\ x+y=56$
the solution $x=30,y=26$ found
the solution of $a,b$ will meet the following equation:
$3a^2+6ab-b^2+1=0\,\, (a,b\in N)$
I wrote a computer program if b<1000000
the corresponding solution of :
[TABLE="width: 183"]
[TR]
[TD="width: 133, bgcolor: transparent"]x=$a^2+ab$

[/TD]
[TD="width: 110, bgcolor: transparent"]y=$ab$

[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]30
[/TD]
[TD="bgcolor: transparent"]26
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]5852
[/TD]
[TD="bgcolor: transparent"]5068
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]1135290
[/TD]
[TD="bgcolor: transparent"]983190
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]220240440
[/TD]
[TD="bgcolor: transparent"]190733816
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]42725510102
[/TD]
[TD="bgcolor: transparent"]37001377138
[/TD]
[/TR]
[/TABLE]
[TABLE="width: 195"]
[TR]
[TD="width: 127, bgcolor: transparent, align: right"][/TD]
[TD="width: 133, bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[/TABLE]
 
Last edited:
kaliprasad said:
show that for x,y positive integers satisfying $3x^2+x= 4y^2+y$ each of x-y , 3x+3y+ 1 and 4x + 4y + 1 are squares. ( above equation has atleast one solution x= 30 and y = 26)

hint

$3x^2+x = 4y^2 + y$
or $3x^2- 3y^2 + x - y = y^2$
or $3(x+y)(x-y) + (x-y) = y^2$
or $(3x+3y+1)(x-y) = y^2$
 
hint 2

$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2$
 
we have $3x^2+x = 4y^2 + y$
or $3x^2- 3y^2 + x - y = y^2$
or $3(x+y)(x-y) + (x-y) = y^2$
or $(3x+3y+1)(x-y) = y^2\cdots\, 1$also

$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2\cdots\,2 $multiply (1) and (2) to get $(x-y)^2(3x+3y+1)(4x+4y+1)=x^2y^2$

or$(3x+3y+1)(4x+4y+1)=(\dfrac{xy}{x-y})^2$so $(3x+3y+1)(4x+4y+1)$ is a perfect squareas $4(3x+3y+1) - 3(4x+4y+1) = 1$ so $(3x+3y+1)$ and $(4x+4y+1)$ are coprime and hence perfect squares and the from (1) or (2) $(x-y)$ is a perfect square
 

Similar threads

MHB Inequality of positive real numbers
  • · Replies 1 ·
Replies
1
Views
1K
MHB [ASK] A Line Intercepting A Circle
  • · Replies 4 ·
Replies
4
Views
1K
MHB How to Simplify Rational Expressions with Variables
  • · Replies 4 ·
Replies
4
Views
2K
High School Solving the Heart Curve Equation: 'y =
  • · Replies 18 ·
Replies
18
Views
4K
MHB Show x³y+y³z+z³x is a constant
  • · Replies 1 ·
Replies
1
Views
1K
MHB Absolute value of real numbers
  • · Replies 1 ·
Replies
1
Views
1K
MHB Solve the equation 8x^2−2xy^2=6y=3x^2+3x^3y^2
  • · Replies 4 ·
Replies
4
Views
2K
MHB Find Value of $x-y+z$ in $x,y,z \in N$
  • · Replies 2 ·
Replies
2
Views
1K
MHB System of Equations: Find Triples $(x,y,z)$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Finding the Equation of a Line Given a Point and Slope
  • · Replies 2 ·
Replies
2
Views
2K