Solution:

Given that s-p > r -q as p < q <= r < s

=> (s-p)^2 > (r-q)^2

=> (s-p)^2 + 4sp > (r-q)^2 + 4rq [becuase sp = rq, we add 4sp in LHS and 4rq in RHS]

=> (s+p)^2 > (r+q)^2

=> s+p >= r+q+1 [as all numbers are integers] -> (1)

Suppose √s - √p = 1 (the other possibility is √s - √p < 1 that we will see later)

=> s + p - 2√sp = 1 => s+p = 1 + 2√qr [becuase sp = rq]

but (1) tells that √qr >= q + r => r = q [By AM-GM rule on positive numbers] and p+s = 2q + 1.

Now ps = q^2 => gcd(p, s) = 1 (the explanation is below)

Because if x divides gcd (p, s) and x is prime (or it will be product of two or more primes, but we assume the base case which covers other case as well), then x would divide q [because p = ax, q = bx => ps = abx^2 = q^2 and gcd(a, x) = 1 and gcd(b, x) = 1) and thus x dividing 2q+1 [= p+s = x(a+b)] is a contradiction => each of p and s is a perfect square [gcd(p, s) = 1].

If √s - √p < 1 then p + s < 1 + 2√ps <= 1 + q + r <= p + s which is a contradiction.

=> In all s and p are perfect squares. Now take X -> only possible s is 81 => p = 64

Now take Y, p > 50 => p can be 64 or 81 but if p = 81 then s = 100 (not possible as s < 100) => p can only be 64. The information on r is required to cross-check if our data in hand is correct and it indeed is as √64.81 = 72.

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