Here Let $4([a,b]+(a,b))^2=t^2$ so we will use $a^{2}b^{2}-t^2=-208$ >>>>>> $(ab-t)(ab+t)=-208$ the rest is easy

no solution

$(a,b)=(4,6)$

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solution : by setting $(a , b) = d ,a = aa_1 , b = bb_1 , [a ,b] = da_1b_1$ , equation becomes $d^4a_1^2b_1^2 + 208 = 4d^2*(a_1b_1 + 1)^2$ which implies that d can be 1,2 or 4 only. equation is further equivalent to : $208 = [2d(a_1b_1 + 1)]^2 - (d^2a_1b_1)^2 = x^2 - y^2$(say) now $208 = x^2 - y^2$ is only possible for $(x,y): (17 , 9) (53 , 51) (28 , 24)$ $x$ must be even , since we have $x = 2d(a_1b_1 + 1)$ so $x = 2d(a_1b_1 + 1) = 28 or 24$ solving for two simultaneous equations : $2d(a_1b_1 + 1) = 28 , d^2(a_1b_1) = 24$ knowing that$ d=2 or 4 $, we obtain $(a_1 , b_1 , d) : (3 , 2 , 2) , (2 , 3 , 2) $ hence , $(a , b): (4 , 6) , (6 , 4)$

Vaibhavmaths wrote: solution : by setting $(a , b) = d ,a = aa_1 , b = bb_1 , [a ,b] = da_1b_1$ , equation becomes $d^4a_1^2b_1^2 + 208 = 4d^2*(a_1b_1 + 1)^2$ which implies that d can be 1,2 or 4 only. equation is further equivalent to : $208 = [2d(a_1b_1 + 1)]^2 - (d^2a_1b_1)^2 = x^2 - y^2$(say) now $208 = x^2 - y^2$ is only possible for $(x,y): (17 , 9) (53 , 51) (28 , 24)$ $x$ must be even , since we have $x = 2d(a_1b_1 + 1)$ so $x = 2d(a_1b_1 + 1) = 28 or 24$ solving for two simultaneous equations : $2d(a_1b_1 + 1) = 28 , d^2(a_1b_1) = 24$ knowing that$ d=2 or 4 $, we obtain $(a_1 , b_1 , d) : (3 , 2 , 2) , (2 , 3 , 2) $ hence , $(a , b): (4 , 6) , (6 , 4)$ Other two $(a , b): (2 , 12) , (12 , 2)$

Just change gcd(a, b) = d, a = xd, b = yd, gcd(x, y) = 1 At this point we have that d^2 divides 208 (you get this after a little bit of multiplying) and the problem becomes trying a couple of possibilities. I really don't want to type the whole solution as I am a bit tired, but you can trust me here, just go this way and you will find everything easily : ) Edit: Oh, sorry, I looked all the posts again and I see that someone wrote the identical solution. Hope you don't mind if I leave this one here : )

(a;b)=(4;6)=(6;4)=(2;12)=(12;2)

Let it be $d=(a, b) \implies$ $a=dx$ and $b=dy$ Where $(x, y) =1$ $[a, b] = \frac{a \cdot b} {d} \implies$ $d^2x^2+d^2y^2+208=4 \cdot ( \frac{xyd^2}{d}+d)^2$ After some calculations we get $208=d^2 \cdot (4x^2y^2+8xy+4-x^2-y^2)$ $208=2^4 \cdot 13$ and $d^2$ is a perfect square. We get the cases $d^2=1,4,16$ We have $(a, b) =(6, 4),(12,2)$