If $a$ and $b$ are natural numbers such that $a^{2} - b^{2}$ is prime number, then $a^{2} + b^{2} - 1$ equals

2 (a + b)

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a - b

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2 a b

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Solution

The correct option is C $2 a b$
Consider $a^{2} - b^{2} = (a + b) (a - b)$ is a prime number
$∴ a^{2} - b^{2}$ is either divisible by $1$ or the number itself.
$∴ a - b = 1$ and $a^{2} - b^{2} = a + b$ $(∵ a + b \neq 1 a s a, b ϵ N)$
On squaring $a^{2} + b^{2} - 2 a b = 1$
$\Rightarrow a^{2} + b^{2} - 1 = 2 a b$
Ans: C

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Similar questions

a

b

a^{2} - b^{2}

a^{2} - b^{2}

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

B = - A^{- 1} B A

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a

b

(a^{2} - b^{2})

(a \neq b)

| a^{2} - b^{2} | = | a^{2} + b^{2} - 2 a b |

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