Responding to a pm:

Ishanvs wrote:

1. if x=y^2 and given that x is integer, then y will always be and integer

No. Given y is an integer, x will be an integer but the other way around is not necessary.

Say x = 5 (an integer) but y is \(\sqrt{5}\) or \(- \sqrt{5}\).

Ishanvs wrote:

2. if we take square root on both sides of the equation x= y^2, we should get √x=|y| ; where √x and y are both integers and where the sign of y is not known

That depends on whether y is an integer. If, say, y = 5/3, x and y are not integers.

Ishanvs wrote:

isn't √P^2=|P| and as we don't know the sign of p .??If that is the case then we get |P|=√q

Yes it is. We don't know the sign of p using stmnt 1 alone. But absolute value of p is equal to \(\sqrt{q}\). So the two will get cancelled off in either case. You might be left with a negative sign but you still get an integer.

Say p = 5

\(p^2 = q = 25\)

\(\sqrt{q} = 5\)

\(p * q / \sqrt{q} = 5 * 25 / 5 = 25\)

Say p = -5

\(p^2 = q = 25\)

\(\sqrt{q} = 5\)

\(p * q / \sqrt{q} = -5 * 25 / 5 = -25\) (still an integer)

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