Diophantine Equation | Integer Solutions?

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[MortezaSabzian-dbsl]

2x²-7xy+3y²=7
This is the acceptable form for us
(ax+by)(cx+dy)=7
Considering that we know what the final form of the equation should be, we leave the coefficients x and y unknown and by multiplying the parentheses, we make it into our equation and determine the coefficients according to what they should, be we do
1) ac=2
2) ad+bc=-7
3) bd=3
We multiply both sides of equation 3 by ad
(ad)²+adbc=-7ad
(ad)²+7ad+6=0
(ad+1)(ad+6)=0
Given that a and b must be integers, we calculate their values

[YahiaNebti]

Hello Syber, I suggest this equation for you :
x^3 + 3x = 756
It's about ideas how to deal with big numbers

[Roq-stone]

I’m lazy so I delve right into the second method first. And 7 is prime so, easy brain work.

[ks]

second way is much easier if you just write 2x^2 - 7xy + 3y^2=7 as
2x^2 - xy + 3y^2 - 6xy = 7, then
x(2x-y) -3y(2x-y) = 7
(2x-y)(x-3y)=7
and then just find values of x and y

[jonathanward]

A few weeks ago Sybermath made the comment that complex numbers are fake. They are not. The real number line is a misnomer. It should be called the physical number line bc of order. The union of the physical number line and the complex number line is reality. We see this in fact with the complex plane.

At the end of every Sybermath video I see his icon and am left with a nagging question: Does Sybermath blow dry?

[RobG]

I appreciate your work.
Have you considered using the method of solving quadratic equations which Po-Shen Loh has been teaching?

[walterwen]

Diophantine Equation: 2x^2 – 7xy + 3y^2 = 7, Integer solutions; x, y = ?
2x^2 – 7xy + 3y^2 = (2x – y)(x – 3y) = 7 = (± 7)(± 1) = (± 1)(± 7)
Let: u = 2x – y, v = x – 3y, 3u – v = 3(2x – y) – (x – 3y) = 5x
(2x – y)(x – 3y) = (u)(v) = (± 7)(± 1) = (± 1)(± 7)
u = ± 7, v = ± 1 or u = ± 1, v = ± 7
1. u = 7, v = 1; 5x = 3u – v = 3(7) – 1, 5x = 20, x = 4, y = 2x – u = 8 – 7 = 1
2. u = – 7, v = – 1; 5x = 3u – v = – 21 + 1 = – 20, x = – 4, y = – 8 + 7 = – 1
3. u = 1, v = 7; 5x = 3u – v = 3 – 7, x = – 4/5; Rejected, not integer
4. u = – 1, v = – 7; 5x = 3u – v = – 3 + 7, x = 4/5; Rejected, not integer
Answer check:
x = 4, y = 1 or x = – 4, y = – 1
2x^2 – 7xy + 3y^2 = (2x – y)(x – 3y) = (8 – 1)(4 – 3) = (– 8 + 1)(– 4 + 3) = 7; Confirmed
Final answer:
x = 4, y = 1 or x = – 4, y = – 1

[gauravraj]

Hii sir.. sir can you post video of geometry that was appeared in British mathematics Olympiad in 2011/12 qno 6 that inequality problem..

[scottleung]

I also used the first method, except I solved for y in terms of x.

[giuseppemalaguti]

L'equazione di 2 grado, in x, da come soluzioni soluzione é per y=1, y=-1..quindi x=4, x=-4.... (altre soluzioni non ne vedo)

[martincohen]

My solution worked out mentally before writing it down.

2x^2-7xy+3y^2=7

d=7^2-4x2x3=49-24=25=5^2
so factors.
Roots are (7+-5)/4=3, 1/2
so 2x^2-7xy+3y^2=(2x-y)(x-3y).

Have 2x-y=a, x-3y=b where
(a, b)=(7, 1), (1, 7), (-7, -1), (-1, -7).

If x, y is a solution, so is -x, -y, so only need to look at x>0.

From 2x-y=a, x-3y=b, have 2x-6y=2b so 5y=a-2b so
y=(a-2b)/5
and 2x=y+a=a+(a-2b)/5=(6a-2b)/5 so
x=(3a-b)/5

a, b=7, 1 gives x=4, y=1.

a, b=1, 7 gives x=-4/5, y=-13/5, not integers, but works.