If p and n are integers such that p n 0 and p^2 - n^2 = 12, which of the following can be the value of p - n?

I. 1
II. 2
III. 4

a. I only
b. II only
c. I and II only
d. II and III only
e. I, II, and III

Question

If p and n are integers such that p > n > 0 and p^2 - n^2 = 12, which of the following can be the value of p - n?

I. 1
II. 2
III. 4

a. I only
b. II only
c. I and II only
d. II and III only
e. I, II, and III

anonymous · Answer

$$(p+n)(p-n) = 12$$

anonymous · Answer

Firstly: $12 = 6 \times 2$

anonymous · Answer

$$(p+n)(p-n) = 6 \times 2$$

anonymous · Answer

If $p=4$ and $n=2$, then is it okay?

aripotta · Answer

wow you're a genius

anonymous · Answer

Wait..

aripotta · Answer

i wouldn't have thought to make it (p + n)(p - n)

anonymous · Answer

Above values make it, so $p-n = 4-2 = 2$

aripotta · Answer

right, so the answer is b?

anonymous · Answer

So, II 2, is right and we have eliminated only a option so far. :(

anonymous · Answer

Also : $12 = 4 \times 3$, with this, we can't find $p$ and $n$ which can be integers.

aripotta · Answer

wait how do you know that? what does making 12 into 4 x 3 do?

anonymous · Answer

12 is product of (6*2), (4*3) and (12*1), right?? Any other set you can make?

aripotta · Answer

no, unless we do -6 and -2, etc

aripotta · Answer

but why do we have to factor the 12?

anonymous · Answer

First one ie (6*2) fits to our equation, as p =4 and n = 2 are integers and they are satisfying our equation, so p-n = 2 is sure..

anonymous · Answer

I am factoring 12, as LHS is also factored..

aripotta · Answer

oh

anonymous · Answer

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