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For how many unique coordinate points $(P,Q)$, such that $P$ and $Q$ are integers, is it true that $P^{2} - Q^{2} = 1155$?

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    Hint : $P^2-Q^2=(P-Q)(P+Q)$. Now factor $1155$ to get the possible pairs. – 2017-02-20
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    Hi Peter, I have factorized 1155 as 3*5*7*11 .... from there I can I get 16 factors of 1155. So it implies that I have 16 possible values to choose p+Q, and the rest is P-Q... alternatively If I chose p-Q first then I would I have another 16 solutions... So, in total it turns out 32.....Is it something like this?? – 2017-02-20
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    Not in general , solve each linear equation system. Only those with integer solutions give a solution. Here, the two factors must be odd, so every system will have an integer solution. Also consider, whether we produce the same pair more than once. – 2017-02-20
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    Okay, Peter, Is it something like I have to firstly consider the positive values and then the negative? So,it turns out 16+16=32... Am I right now Peter? – 2017-02-20
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    There is no way to get the same pair more than once as the both factors are prime...am I right peter? – 2017-02-20
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    I am not able to answer this without brute force. Maybe there is a proof that the pairs are all distinct, but I do not know it. You mean : coprime ? – 2017-02-20
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    I wrote a small program and $32$ is actually correct. – 2017-02-20
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    Thank you, Peter, for your effort.... But how do people solve this type of problem by 2 minutes at the GRE test? – 2017-02-20
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    Your thoughts are correct, I assume they simply followed them and simply counted the factors , which is not difficult here. Nevertheless, $2$ minutes is a good performance! – 2017-02-20

1 Answers 1

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HINT: $$P^2-Q^2=1155=11\times 105=3\times 5 \times 7\times 11$$ $$(P+Q)(P-Q)=3\cdot 5\cdot 7\cdot 11$$

Now, if the RHS is grouped to represent it as the product of $2$ numbers only, then in how many ways can you make such groups? The $4$ numbers on the RHS are mutually prime, hence any such choice will yield a unique pair.

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    It is OK now @SchrodingersCat – 2017-02-20
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    @DonaldSplutterwit Oh thanks... It was a typo. – 2017-02-20
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    @Rohan Thanks for the edit. – 2017-02-20
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    @SchrodingersCat No problem. – 2017-02-20