A number theory problem by U Z

$n + 9 = a^{2}$

16n + 9 = $b^{2}$

27n + 9 = $c^{2}$

$16a^{2}$ = 16n + 144

$b^{2}$ = 16n + 9

then

$16a^{2} - b^{2}$ = 135

(4a - b)(4a +b) =135 = 1X45, 3X45, 5X27, 9X15

a = 17,6,4,3

n= 280 ,27,7,0

27n + 9 = 9(3n +1) = 9(841),9(22),9(1)

9(841) = $87^{2}$

when n = 280

n + 9 = 289 = $17^{2}$

16n + 9 = $67^{2}$

27n + 9 = $87^{2}$

therefore only one solution

simply,, i guess the answer 1.. i thought i should start putting the value of n from 1.. n i noticed that if n=1 then, n+9 = 9 is a perfect square. Similarly, if n=1 then, 16n+9 = 25 is a perfect square n so on..!! simple as that :v ;)