If P1 and P2 are two odd prime numbers such that P1 > P2 , then P12 -P

1.	If P1 and P2 are two odd prime numbers such that P1 > P2 , then P12 -P22 is …………….. A) an even numberB) an odd numberC) a prime numberD) an odd prime number
Answer» Correct option is (A) an even number Given that \(P_1\) and \(P_2\) are two odd numbers such that \(P_1>P_2.\) Let \(P_1\) = 2m+1 and \(P_2\) = 2n+1 Then m > n \((\because P_1>P_2)\) Now, \(P_1^2-P_2^2\) \(=(2m+1)^2-(2n+1)^2\) \(=(4m^2+4m+1)-(4n^2+4n+1)\) \(=4(m^2-n^2)+4(m-n)\) = 4 (m-n) (m+n+1) \(\therefore\) 4 divides \(P_1^2-P_2^2\) \(\therefore\) 2 divides \(P_1^2-P_2^2\) Thus, \(P_1^2-P_2^2\) is an even number. Correct option is A) an even number

If P1 and P2 are two odd prime numbers such that P1 > P2 , then P12 -P22 is …………….. A) an even numberB) an odd numberC) a prime numberD) an odd prime number

Answer»

Correct option is (A) an even number

Given that \(P_1\) and \(P_2\) are two odd numbers such that \(P_1>P_2.\)

Let \(P_1\) = 2m+1 and

\(P_2\) = 2n+1

Then m > n \((\because P_1>P_2)\)

Now, \(P_1^2-P_2^2\) \(=(2m+1)^2-(2n+1)^2\)

\(=(4m^2+4m+1)-(4n^2+4n+1)\)

\(=4(m^2-n^2)+4(m-n)\)

= 4 (m-n) (m+n+1)

\(\therefore\) 4 divides \(P_1^2-P_2^2\)

\(\therefore\) 2 divides \(P_1^2-P_2^2\)

Thus, \(P_1^2-P_2^2\) is an even number.

Correct option is A) an even number