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If \(p^2\) is an integer and \(\sqrt{p^6-p^4-q-1}=10\) what is the value of \(p^2\)?


(1) \(p^2=\sqrt{p^2+20}\)

(2) \(q=\sqrt{q+2}\)


(1) \(p^2=\sqrt{p^2+20}\)
Square both sides..
\(p^4=p^2+20............p^4-p^2-20=0.........(p^2-5)(p^2+4)=0\)
As \(p^2\) is a square root it has to be a positive integer. Also, If \(p^2 = -4\), then p will be an imaginary number.
Thus \(p^2=5\)
Sufficient

(2) \(q=\sqrt{q+2}\)
Square both sides..
\(q^2=q+2......q^2-q-2=0....(q+1)(q-2)=0\)
q can be 2 or -1, but again q can be positive so 2.
Putting 2 in the equation \(\sqrt{p^6-p^4-q-1}=10\) does not give p^2 as 5. So, I believe the statement should be
\(q=-\sqrt{q+2}\)
So q is negative, and so q=-1.
Put this value in \(\sqrt{p^6-p^4-q-1}=10\)
\(\sqrt{p^6-p^4-(-1)-1}=10\)
\(\sqrt{p^6-p^4}=10\)
\(p^6-p^4=100\)
Three possible values of p^2, but only integer value is 5, and other two are imaginary values.
Sufficient

D

Please check statement II for typo.
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