If a and b are natural numbers such that 2013 | vedclass

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(C) Given the equation $2013 + a^2 = b^2$.Rearranging the terms,we get $b^2 - a^2 = 2013$.Using the difference of squares identity,$(b - a)(b + a) = 2

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$658$

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(C) Given the equation $2013 + a^2 = b^2$.Rearranging the terms,we get $b^2 - a^2 = 2013$.Using the difference of squares identity,$(b - a)(b + a) = 2013$.The prime factorization of $2013$ is $3 	imes 11 	imes 61 = 33 	imes 61$.Since $a$ and $b$ are natural numbers,$b + a > b - a$ and both must be positive factors of $2013$.To minimize $ab$,we look for factors $(b - a)$ and $(b + a)$ that are closest to each other.Let $b - a = 33$ and $b + a = 61$.Adding the two equations: $2b = 94 \Rightarrow b = 47$.Subtracting the two equations: $2a = 28 \Rightarrow a = 14$.Thus,the minimum value of $ab = 14 	imes 47 = 658$.

If $a$ and $b$ are natural numbers such that $2013 + a^2 = b^2$,then the minimum possible value of $ab$ is:

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