Math, asked by SinISteR7253, 9 months ago

If p1 and p2 are two odd prime numbers such that p1 ›p2 then p1^2 - p2. ^2 is

Answers

Answered by amitnrw
13

If p1 and p2 are two odd prime numbers such that p1 ›p2 then p1^2 - p2 is always even number and also always divisible by 8

Step-by-step explanation:

Let say p1 = 2n + 3    & p2 = 2n + 1

p1²  - p2²

= (2n + 3)²  - (2n + 1)²

= (4n² + 9 + 12n) - (4n² + 1 + 4n)

= 8n + 8

= 8(n + 1)

= Even number  ( Always Divisible by 8)

If p1 and p2 are two odd prime numbers such that p1 ›p2 then p1^2 - p2 is always even number and also always divisible by 8

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Answered by Anonymous
9

Answer: Even number

Step-by-step explanation:

Let p1 = 2n+1 and p2=2n-1. These are odd numbers, for any odd number is given by 2n+1 or 2n-1.

Therefore, p12-p22 = (p1+p2) (p1-p2), as you know that a2-b2 = (a+b) (a-b)

Substituting the values of p1 and p2 in the above equation, we get

p12-p22 = (2n+1+2n-1) (2n+1-2n+1) = 4n * 2 = 8n

Thus, p12-p22 is always even, for any n.

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