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

  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Numbers Square
Click For Summary
The equation $3x^2 + x = 4y^2 + y$ has positive integer solutions, with one known solution being x = 30 and y = 26. For these values, it is demonstrated that the expressions x - y, 3x + 3y + 1, and 4x + 4y + 1 are all perfect squares. The discussion emphasizes the need to prove that these expressions yield square numbers for any positive integer solutions to the equation. The mathematical relationships and properties of squares are central to the analysis. This exploration contributes to a deeper understanding of the equation's solutions and their implications.
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
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K