Editorial for A Square Problem

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Author: Ninjaclasher

Note that $a^2-b^2$ ~a^2-b^2~ is a difference of squares. This means it can be rewritten as $(a-b)(a+b)$ ~(a-b)(a+b)~.

From this, it can be proven that $a^2-b^2$ ~a^2-b^2~ is only prime when $a-b=1$ ~a-b=1~ and $a+b$ ~a+b~ is a prime.

Thus, we can check if $a+b$ ~a+b~ is prime to determine if $a^2-b^2$ ~a^2-b^2~ is prime. To check if $a+b$ ~a+b~ is prime, we can check if it is divisible by any numbers less than its square root. The proof is left as an exercise for the reader.

Time Complexity: $\mathcal{O}(\sqrt{a+b})$ ~\mathcal{O}(\sqrt{a+b})~

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