WLOG $x \leq y$. Let $(x,y)=d$, $x=ds$, $y=dt$, so $[x,y]=dst$. Thus we must find positive integer solutions to $\frac{1}{ds}+\frac{1}{dt}+\frac{1}{d}+\frac{1}{dst}=\frac{1}{2} \iff \frac{s+t+1}{st} = \frac{d-2}{2}$. Note that $d \geq 3$. If $d \geq 9$, we have $\frac{s+t+1}{st} \geq \frac{7}{2}$, so $7st - 2s - 2t - 2 \leq 0 \iff 25 \leq (7s-2)(7t-2) \leq 18$, which is a contradiction. Therefore, $3 \leq d \leq 8$. $$\frac{s+t+1}{st} = \frac{d-2}{2} \iff ((d-2)s-2)((d-2)t-2)=2d$$If $d=3$, we have solutions for $(s,t)$ as $(3,8)$ and $(4,5)$. If $d=4$, we have solutions for $(s,t)$ as only $(2,3)$. If $d=5$, we have solutions for $(s,t)$ as $(1,4)$. If $d=6$, we have solutions for $(s,t)$ as $(1,2)$. If $d=7$, we have no solutions for $(s,t)$. If $d=8$, we have solutions for $(s,t)$ as $(1,1)$. These give solutions for $(x,y)$ as $(5, 20), (6, 12), (8, 8), (8, 12), (9, 24), (12, 15)$. Therefore, all possible solutions for $(x,y)$ are $(5, 20), (6, 12), (8, 8), (8, 12), (9, 24), (12, 6), (12, 8), (12, 15), (15, 12), (20, 5), (24, 9)$.

I am sharing a longer,but easier solution: Let $x=d_1k$,$y=d_2k$ where $(d_1,d_2)=1$.Equation is equivalent to $2(d_1+1)(d_2+1)=d_1d_2k$.$(d_1;d_1+1) =(d_2;d_2+1)=1$,so,we must have $d_1|2(d_2+1)$ and $d_2|2(d_1+1)$.Let $d_2=\frac{d_1x-2}{2}$.Putting this in the second equation as an inequality gives $d_1x-2\leq4d_1+4$.$x\geq11$ gives $7d_1\leq6$,contradiction.We must check $x\leq10$. 1)$x=10$ gives:$d_1=1$,so,$d_2=4$ and $k=5$.So,(5;20),(20;5) are solutions. 2)No solution out of $x=9$. 3)$x=8$ gives $d_1=1$,$d_2=3$.Contradiction due to original equation. 4)$x=7$ gives $d_1=2$,$d_2=6$,$(d_1,d_2)=2$,contradiction 5)$x=6$ gives $d_1=1,2,3$ and $d_2=2,5,8$ for corresponding values.Checking these values gives $d_1=1,d_2=2,k=6\rightarrow$ (6;12),(12;6) are solutions.$d_1=3,d_2=8,k=3\rightarrow$ (9;24),(24;9) are solutions as well. 6)$x=5$ gives no solutions. 7)$x=4$ is equivalent to $d_2=2d_1-1$.Putting this in the original equation gives $4d_1+4=k(2d-1)\rightarrow2d_1-1|4d_1+4\Rightarrow2d_1-1|6$.We get $d_1=1=d_2\rightarrow k=8$ (8;8) is a solution.Or,$d_1=2,d_2=3,k=4\rightarrow$ (8;12),(12;8) is a solution. 8)$x=3$ gives $3d_1-2|6d_1+6\Rightarrow3d_1-2|10$.We only get $d_1=d_2=1$ which we solved before as well. 9)$x=2$ gives $d_1-1|2d_1+2\Rightarrow d_1-1|4$.$d_1=2,3$ has been solved before.We look at $d_1=5,d_2=4,k=3$ which gives solutions (15;12),(12;15). 10)$x=1$ doesn't give any new solution.

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Sadigly wrote: Like count is just a number : P Post count is just a number : P

Nuran2010 wrote: Sadigly wrote: Like count is just a number : P Post count is just a number : P Bruh I have around 200 posts which is not counted idgaf about post numbers