Sorry not sure if a post of the same problem already there...
Let , then
is a perfect square if and only if are all perfect squares.
The statement is wrong. It is not an "if and only if" condition. Take , the two first factors are not perfect squares but the expression is a perfect square.
Anyway, assuming it is an "if" condition (if the three factors are perfect square then the expression is a perfect square) it is easy to show by using the property that the product of two or more squares is a square.
If you're studying in some (more ore less decent) university then they either have the article in question in the maths library or else they've a subscription to Jstore. Use them.
The theorem being proved in that paper doesn't look hard but it does seem to be way too involved in all kinds of calculations.
Tonio