Fool's problem of the day,
How many ordered pair of \((x,y)\) are there such that \(x, y \in \mathbb{Z} \) and \( \frac 2 x - \frac 3 y = \frac 1 4 \)
Good luck!
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Fool's problem of the day,
How many ordered pair of \((x,y)\) are there such that \(x, y \in \mathbb{Z} \) and \( \frac 2 x - \frac 3 y = \frac 1 4 \)
Good luck!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
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okay, we can get after simplification..
x=8y/(y+12)
as x is an integer, 8y must be a multiple of y+12
8y=2*2*2*y
so, here are possible equations
y+12=1
y+12=2
y+12=4
y+12=8
y+12=2y
y+12=4y
y+12=8y
as y +12 is a factor of 8y
all the equations except the last one will satisfy for an integral value of y
so, the answer is.. 6 ordered pairs
sorry, there will be other options too if RHS is negative
y+12=-1
y+12=-2
................. and so on..
for y to be an integer, there are another 5 solutions
so, total 11
I was able to find 11 ordered pairs that are solutions:
(7, 84)
(6,36)
(5,20)
(4,12)
(2,4)
(-4,-4)
(-8,-6)
(-16,-8)
(-24,-9)
(-40,-10)
(-88,-11)
I think there might be more, but im missing an idea.