x/y+17/4 is a positive integer.
=> x/y = 3/4+t, where t is a nonnegative integer.
=> x/y= (3+4t)/4
Then, x/(3+4t) =y/4 = k(say)
=> x = (3+4t)*k and: y =4k ....... (*)
Then, we must have:
k^2((3+4t)^2 + 4^2) = {(3+4t+17)/4}^2
=> k^2{(3+4t)^2+4^2}=(5+t)^2
Thus, (3+4t)^2+4^2 must be a perfect square
So, (3+4t)^2+4^2= m^2 (say)
=> (m+3+4t)(m-3-4t) = 4^2=16
=> (m+3+4tm m-3-4t) = (16,1), (8,2)
Now, the pair (m-3-4t, m-3+4t) =(16,1) does NOT lead to an integer solution.
(m-3-4t, m-3+4t) = (8,2) gives:
=> (m, 3+4t) = (5,3)
=> (m,t) = (5,0)
Then, for t=0, we have from (*):
x=3k, y =4k
Then, we have:
(3k)^2+(4k)^2 = (3/4+17/4)^2
=> 5^2 * k^2 = 5^2
=> k^2 =1
=> k =1 (disregarding the value of -1, which is NOT positive.)
Then, substituting t=0 and k =1 in (*) gives:
(x,y) = (3,4) which is the only possible solution to the given problem.
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