Equation with several variables

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The equation ## x^2 - y^2 - x + 3y = 30 ## can be transformed into the product form ## (x - y + 1)(x + y - 2) = 28 ##. By defining ## u = x - y + 1 ## and ## v = x + y - 2 ##, it is established that both ## u ## and ## v ## must be divisors of 28, sharing the same sign. The valid pairs of ## (u, v) ## lead to the integer solutions for ## (x, y) ##, which include the points { (-14, -12), (15, 15), (-5, 0), (6, 3), (-5, 3), (6, 0), (-14, 15), (15, -12) }. A correction was noted regarding the inclusion of ## \pm 14 ## in the divisors of 28.

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Homework Statement


Solve in ##\mathbb{Z}^2## the equation ## x^2 -y^2-x+3y = 30 ##

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The Attempt at a Solution



Hello, can you tell me if this is correct please ?

The equation is equivalent to ## (x-y+1) (x+y-2) = 28 ##.
I call ## u = x-y+1 ## and ## v = x+y-2 ##
We have that ## u | 28 ##, ## v | 28 ##, and ##uv = 28##

So, ##u,v \in \text{Div}(28)=\{\pm 1, \pm 2, \pm 4, \pm 7, \pm 28 \} ## and share their sign

Finally, we must have ##(u,v) \in## { (-1,-28) , (1,28), (-2,-14), (2,14), (-4,-7) ,(4,7), (-7,-4), (7,4), (-14,-2), (14,2), (-28,-1),(28,1) }

So ## (x,y) = ( \frac{u+v+1}{2}, \frac{v-u+3}{2} ) \in ## { (-14,-12),(15,15),(-5,0),(6,3),(-5,3), (6,0), (-14,15), (15,-12) }
 
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Thanks, there was a typo, I forgot ##\pm 14## in Div(28) but I took it into account
 

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