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# Is (p - 2)(q - 2) square of an integer?

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Re: Is (p - 2)(q - 2) square of an integer? [#permalink]
Expert Reply
kiran120680
Is (p-2) (q-2) square of an integer?

(1) pq = 2(p+q)

(2) p – q = 0
Solution:
Pre Analysis:
• We are asked if $$(p-2) (q-2)$$ is square of an integer or not
• $$(p-2) (q-2)=pq-2p-2q+4=pq-2(p+q)+4$$
• We are asked if $$pq-2(p+q)+4$$ is square of an integer or not

Statement 1: $$pq = 2(p+q)$$
• $$pq-2(p+q)+4$$
$$=2(p+q)-2(p+q)+4$$
$$=4$$
• We know 4 is square of integer 2
• Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: $$p – q = 0$$
• $$p-q=0$$ means $$p=q$$
• $$pq-2(p+q)+4$$
$$=p^2-2(p+p)+4$$
$$=p^2-4p+4$$
$$=p^2-4(p-1)$$
• We are not sure if $$p^2-4(p-1)$$ is square of an integer or not
• Thus, statement 2 alone is not sufficient

Hence the right answer is Option A
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Re: Is (p - 2)(q - 2) square of an integer? [#permalink]
Cutemon
Statement 1

pq = 2(p+q)
So for only value of 4 for both p and q, the condition satisfy.
i.e. 4*4 = 2(4+4)
16 = 16
Statement 1 is sufficient.

Statement 2:
p - q = 0
This implies that p and q are of same value and same sign.There can be any range of values for p and q, i.e. (2,2), (4,4), (100,100).
But for p and q having value 2, the condition (p-2)(q-2)=square of an integer doesn't satisfy, as 0*0 is not a perfect square.
Statement 2 is insufficient condition.

Answer is A

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­I would argue with this, as the question is not if (p-2)(q-2) is a perfect square (defined as the product of two positive integers), but simply if it is the quare of an integer (nowhere defined if positive or anything alike). And the last time I checked, 0 is a an integer and 0 is also the square of 0 (0*0=0)...
Re: Is (p - 2)(q - 2) square of an integer? [#permalink]
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