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The answer is part $d)$none of the above, How do i proceed with such questions?

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    $(p^2-q^2)=(p-q)(p+q)$, from the fact that $p^2-q^2$ is a prime number and $p+q>p-q$, what do you have? – 2017-01-24
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    It can't be a prime number unless p-q=1 – 2017-01-24
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    I know you know how to use MathJax, so you should take the time to write out what the textbook or handout says instead of just posting a photo of it. – 2017-01-25

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Note that $$p^2-q^2=(p-q)(p+q)$$

From the definition of prime numbers, it follows that $p-q=1$.

If $p-q \neq 1$, then $p-q$ is a number that divides $p^2-q^2$ that is not $1$ or $p^2-q^2$.

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    so $p=3$ and $q=2$ – 2017-01-24
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    @JorgeFernándezHidalgo I'm afraid I don't see how that follows. – 2017-01-24
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    @JorgeFernándezHidalgo On the contrary, $p=4, q=3$ are solutions as well. There is no mention that $p,q$ are prime numbers. – 2017-01-24
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    oh, my bad. I assumed by the variables – 2017-01-24
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    And if $p-q$ is not$ 1$ then $p^2-q^2$ ceases to be a prime number. I understand your solution, Thankyou! – 2017-01-24
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    @MAthsjunky Your welcome. :) Would you accept my solution? – 2017-01-24
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    why can't it be -1? – 2017-01-24
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    @SubhashChandBhoria A prime number is a natural. If $p-q=-1$, $p+q>0$, then $p^2-q^2$ is a negative number. So it can't be a prime number. – 2017-01-24
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given Since $p,q$ are integers and $p^2-q^2=(p-q)(p+q)$ is prime, it means that we got integer factors of primes number. And since we know that the only factors of prime are $1$ and prime only therefore we are force to believe that $p-q=+1$ or $-1$ which leaves no other choice as integrs are positive