JEST Sample Question 1-a

Question

Let a and b be positive integers such that a > b and a^ 2 − b^ 2 is a prime number.
Then a^2 − b^ 2 is equal to
(A) a − b
(B) a + b
(C) a × b
(D) none of the above

Comments

option c is right answer -

we can try it with the help of some inputs

take a=3,b=2 then $3^2-2^2=5(prime ) $ and which is equal to a+b

take another a=4,b=3 then $4^2-3^2=7$ and which is equal to a+b

Answer 1

For conditions: $a>0, b>0, a>b$ & $a^2$ − $b^2$ to be prime number

=> a & b must be consecutive integer i.e., $(a-b)$ = $1$        (for explanation see edit)

thus, $a^2$ − $b^2$ = $(a-b)*(a+b)$ = $(a+b)$.

Hence, option $(B)$ is correct.

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EDIT: factors of prime number: that number itself & 1.

since ($a^2$-$b^2$) have two factors $(a+b)$ & $(a−b)$ and if $(a-b)$ will be any number other than 1 than with given conditions $[a>b,a>0,b>0]$ , ($a^2$-$b^2$) will never be prime number.

so only choice is $(a−b)$ = 1.

Comments

 

=> a & b must be consecutive integer i.e., (a−b) = 1

how you conclude this ?

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factors of prime number: that number itself & 1.

since $a^2$-$b^2$ have two factors $(a+b)$ & $(a-b)$ and if (a-b) will be any number other than 1 than with given conditions $[a>b, a>0,b>0]$ , ($a^2$-$b^2$) will never be prime number.

so only choice for $(a-b)$ = 1.

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keep this info in the answer itself !