well..write this like$(2x-3y)^2+(4p^2-9)y^2=48p$..Separate cases..first with $y=1$ which gives $153-4(p-6)^2$ should be a square..no sol in this case(i checked roughly,.but i thnk its r8)..second case..$y\geq 2$..which gives $48p\geq 4(4p^2-9)$ or $4p^2-12p-9\leq 0$ equivalent to $(2p-3)^2\leq 18<25$ which gives $p<4$.so $p=2,3$..check thiese..$p=2$ gives $(2x-3y)^2+7y^2=96$ check $y=1,2,3$ is only possible but gives no solution.. $p=3$..well the equation is like $(2x-3y)^2+27y^2=144$..only possiblity $y=1,2$..$y=2$ gives $x=6$..So only sol is $(6,2,3)$...well again saying checked roughly..there may be some mistakes..but the procedure is same..

Observe that $x^2+p^2y^2 \vdots 3 \Rightarrow x \vdots 3, py \vdots 3$ so the LHS must be divisible by 9, hence the RHS must also divisible by 9, thus we have $p=3$. Plug in and solve standard quadratic diophantine equation, we have $(x,y)=(0,2),(0,-2),(6,2),(-6,-2)$

note that , for integral $x$ , we must have , $y^2(9-4p^2)+48p=z^2$ note that , if $p \neq 2,3$ , then 3 divides $x^2+y^2$ , so , 3 divides x and y so , 9 divides $[z^2-3p]$. if $p=1(mod3)$ , then 9 divides $z^2-3$ , which is impossible. similarly , we get a contradiction for $p=-1(mod3)$. so , we need to check p=2,3 only which is trivial.

this question was in my exam in Azerbaijan.....

crazyfehmy wrote: Let $x, y$ be integers and $p$ be a prime for which \[ x^2-3xy+p^2y^2=12p \] Find all triples $(x,y,p)$. The equation can be rewrite as $x(x - 3y) = p(12 - py^2)$. if $p = xd$ then $d(pd - 3y) = 12 - py^2$ $\implies$ $p(d^2+y^2) = 3(4 + yd)$. If $p = 3$ then $d^2 - yd + y^2 - 4 = 0$ so we get that $16 - 3y^2$ is a perfect square so $y = 2$ or $y = -2$ then $d \in {0,2}$ so $(x,y,p) \in \{(0,2,3),(6,2,3),(0,-2,3),(6,-2,3)\}$. if $p$ isn't $3$ then $d,y \equiv 0 ($ $mod$ $3)$. then $4 \equiv 0 (mod$ $3)$,contradiction. We do similar when $x - 3y = pd$.

Doesn't the triple $(6,0,3)$ work?

Rewrite as following $x^2-3xy+p^2y^2-12p=0$ Clearly the discriminant $9y^2-4p^2y^2+48p = d^2$ should be a perfect square. If $p\neq3$ then $p^2\equiv 1(mod$ $3)$ and $9y^2-4p^2y^2+48p\equiv -4p^2y^2\equiv 2y^2 \equiv d^2(mod$ $3)$ Since quadratic residues $(mod$ $3)$ are $1$ and $0$, we can conclude that $3|y$ $\Rightarrow$ $y=3y_1$ $x^2-9xy_1+9p^2y_1^2=12p$ $\Rightarrow$ $3|x$ $\Rightarrow$ $x=3x_1$ $9x_1^2-27x_1y_1+9p^2y_1^2=12p$ $3x_1^2-9x_1y_1+3p^2y_1^2=4p$ $3|LHS$ but not the right side, hence contradiction. $\Rightarrow$ $p=3$ $x^2-3xy+9y^2=36$ $\Rightarrow 3|x$ $\Rightarrow x=3x_1$ $x_1^2-x_1y+y^2=4$ We solve this quadratic equation by first checking the discriminant and making it be a perfect square and we get the following solutions $(p,x,y)=(3,6,0),(3,6,2),(3,0,2),(3,0,-2),(3,-6,-2)$

harapan57 wrote: Rewrite as following $x^2-3xy+p^2y^2-12p=0$ Clearly the discriminant $9y^2-4p^2y^2+48p = d^2$ should be a perfect square. If $p\neq3$ then $p^2\equiv 1(mod$ $3)$ and $9y^2-4p^2y^2+48p\equiv -4p^2y^2\equiv 2y^2 \equiv d^2(mod$ $3)$ Since quadratic residues $(mod$ $3)$ are $1$ and $0$, we can conclude that $3|y$ $\Rightarrow$ $y=3y_1$ $x^2-9xy_1+9p^2y_1^2=12p$ $\Rightarrow$ $3|x$ $\Rightarrow$ $x=3x_1$ $9x_1^2-27x_1y_1+9p^2y_1^2=12p$ $3x_1^2-9x_1y_1+3p^2y_1^2=4p$ $3|LHS$ but not the right side, hence contradiction. $\Rightarrow$ $p=3$ $x^2-3xy+9y^2=36$ $\Rightarrow 3|x$ $\Rightarrow x=3x_1$ $x_1^2-x_1y+y^2=4$ We solve this quadratic equation by first checking the discriminant and making it be a perfect square and we get the following solutions $(p,x,y)=(3,6,0),(3,6,2),(3,0,2),(3,0,-2),(3,-6,-2)$ Nice solution harapan!

harapan57 wrote: Rewrite as following $x^2-3xy+p^2y^2-12p=0$ Clearly the discriminant $9y^2-4p^2y^2+48p = d^2$ should be a perfect square. If $p\neq3$ then $p^2\equiv 1(mod$ $3)$ and $9y^2-4p^2y^2+48p\equiv -4p^2y^2\equiv 2y^2 \equiv d^2(mod$ $3)$ Since quadratic residues $(mod$ $3)$ are $1$ and $0$, we can conclude that $3|y$ $\Rightarrow$ $y=3y_1$ $x^2-9xy_1+9p^2y_1^2=12p$ $\Rightarrow$ $3|x$ $\Rightarrow$ $x=3x_1$ $9x_1^2-27x_1y_1+9p^2y_1^2=12p$ $3x_1^2-9x_1y_1+3p^2y_1^2=4p$ $3|LHS$ but not the right side, hence contradiction. $\Rightarrow$ $p=3$ $x^2-3xy+9y^2=36$ $\Rightarrow 3|x$ $\Rightarrow x=3x_1$ $x_1^2-x_1y+y^2=4$ We solve this quadratic equation by first checking the discriminant and making it be a perfect square and we get the following solutions $(p,x,y)=(3,6,0),(3,6,2),(3,0,2),(3,0,-2),(3,-6,-2)$ You forgot (3,-6,0)

$x^2=12p+3xy-p^2y^2\equiv{-p^2y^2}(mod 3)$ Since $p^2y^2\equiv{0 ; 1 }(mod 3)$ So $x^2\equiv{0}(mod 3)$ $\rightarrow{x\equiv{0}(mod 3)}$ $\rightarrow{p^2y^2\equiv{0}(mod 3)}$ Hence we get 2 cases such that : $(i)$ $y\equiv{0}(mod 3)$ Then $LHS\equiv{0}(mod 9)$ $\Rightarrow{12p\equiv{0}(mod 9)\rightarrow{p=3}}$ $(ii)$ $p\equiv{0}(mod 3)$ implies the same thing that $p=3$ is the only solution . So the equation becomes : $x^2-3xy+9y^2=36$ $\Rightarrow{9y^2-3xy+(x^2-36)=0}$ $\Delta=9x^2-36(x^2-36)>0$ Which implies that $36^2>27x^2\rightarrow{48>x^2}$ Don't forget that $x\equiv{0}(mod 3)$ So only candidates are $x\in{(0 ; 3 ; 6 ; -3 ; -6)}$ Testing we see that the solutions are : $(x;y;p)=(6;0;3)=(-6;0;3)=(0;2;3)=(0;-2;3)=(-6;-2;3)=(6;2;3)$ .$Q.E.D$

Nguyenhuyhoang wrote: Observe that $x^2+p^2y^2 \vdots 3 \Rightarrow x \vdots 3, py \vdots 3$ so the LHS must be divisible by 9, hence the RHS must also divisible by 9, thus we have $p=3$. Plug in and solve standard quadratic diophantine equation, we have $(x,y)=(0,2),(0,-2),(6,2),(-6,-2)$ Good solution bravo easiest in this Propblem l can help to solve quadratic equation Solution:After puting p=3 We get x²-3xy+(9y²-36)=0 Here ∆=144-27y² Here 144-27y²must be perfect square for x integer and only y here is y=2,-2 Puting it we found x by ∆ and then we will get the Answer