Answer:
This statement can be proven by contradiction for 
(including the case where 
.)
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Step-by-step explanation:
Assume that the natural number is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number 
(
) such that 
.
Assume by contradiction that 
is indeed a perfect square. Then there should exist another natural number 
such that 
.
Note, that since 
, 
. Since 
while 
, one can conclude that 
.
Keep in mind that both and 
are natural numbers. The minimum separation between two natural numbers is 
. In other words, if , then it must be true that 
.
Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by 
and use the fact that to make the left-hand side 
.)
.
Expand the right-hand side using the binomial theorem:
.
.
However, recall that it was assumed that and 
. Therefore,
.
.
Subtract 
from both sides of the inequality:
.
.
Recall that was assumed to be a natural number. In other words, 
and must be an integer. Hence, the only possible value of would be 
.
Since could be equal , there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that 
just won't work as in the assumption.
If indeed , then 
. 
, which isn't a perfect square. That contradicts the assumption that if is a perfect square, would be a perfect square. Hence, by contradiction, one can conclude that
.
Note that to produce a more well-rounded proof, it would likely be helpful to go back to the beginning of the proof, and show that 
. Then one can assume without loss of generality that . In that case, the fact that 
is good enough to count as a contradiction.
