How to solve this diophantine equation :

Can I use quadratic residue here? How can I use it??

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