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

  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Numbers Square
AI Thread 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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
1
Views
1K
Replies
4
Views
2K
Replies
4
Views
1K
Replies
18
Views
4K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
1
Views
1K
Back
Top