How to solve this diophantine equation :
x^2-12y+4=0
Can I use quadratic residue here? How can I use it??
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This is not hard until you grasp the penultimate step,
(Quadratic residue is not always helpful in diophantine equations.)
x^2-12y+4=0
→ -x^2+12y=4
→ (\sqrt{12y})^2-x^2=4
→ (\sqrt{12y}+x)(\sqrt{12y}-x)=4 .... (1)
Now recall the multiples of 4, they are-1,4,2
Hence, from (1), there are 3 pairs of possibilities-
(1) {\sqrt{12y}+x=4, \sqrt{12y}-x=1}
(2) {\sqrt{12y}+x=1, \sqrt{12y}-x=4}
(3) {\sqrt{12y}+x=2, \sqrt{12y}-x=2}
Solving these, you’ll get :-
(x,y)=(\frac{3}{2},\frac{25}{48}),(\frac{-3}{2},\frac{25}{48}),(0,\frac{1}{3}).
So, \fbox{There is no integer solution to the equation}
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Oh, sorry this problem can also be done with quadratic residue.
Just take mod 12…
I couldn’t find it before…
How to do it?? with mod 12
… x^2-12y+4=0
→ x^2=12y-4
→ x^2=4(3y-1)
→ x^2=4u^2
… u^2=3y-1
→ u^2\equiv 2\pmod{3}
But this is impossible, because the quadratic residues of 3 are 0,1,1.
I haven’t worked with it, I just asked it in quora.
x^2-12y+4=0
x^2+4=12y
So x^2+4 should be divisible by 12…
x^2\equiv 0,1,4,9\left(\mod 12\right)
Then, 0+4≠ (0,12)
1+4≠(0,12)
4+4≠(0,12)
9+4≠ (0,12)
If x^2+4 is divisible by 12, after we add the remainders it would be 12…so It does not have solutuons…That was easy peasy
Start by rearranging:
x^2 - 12y + 4 = 0
or, x^2 = 12y - 4
or, x^2 = 4(3y-1)
Notice, 4(3y-1) must be perfect square. 4 is also a perfect square. Thus, 3y-1 is a perfect squares. Let z be the the square root of 3y-1
Notice, 3y-1 \equiv 2 (\bmod 3)
If, z \equiv 0 (\bmod 3), 3y-1 \equiv 0 (\bmod 3) {Contradiction}
If, z \equiv 1 (\bmod 3), 3y-1 \equiv 1 (\bmod 3) {Contradiction}
If, z \equiv 2 (\bmod 3), 3y-1 \equiv 1 (\bmod 3) {Contradiction}
Thus, there is no integer solution for (x,y)
Additional note: I’m pretty sure this is taken from 2023 National BDMO junior problem set. In the og question, it asked for integer solutions only.