Question 13 of 26intermediate🔧 ApplyLong Answer3 marks

If p≥5p \geq 5 is a prime number, show that p2+2p^2 + 2 is divisible by 3.

Correct Answer

Since p≥5p \geq 5 is a prime number, p>3p > 3. By Example 11, every prime number greater than 3 is of the form 6k+16k + 1 or 6k+56k + 5, where kk is some integer.

Case 1: Let p=6k+1p = 6k + 1.

p2=(6k+1)2=36k2+12k+1p^2 = (6k + 1)^2 = 36k^2 + 12k + 1 p2+2=36k2+12k+3=3(12k2+4k+1)p^2 + 2 = 36k^2 + 12k + 3 = 3(12k^2 + 4k + 1)

which is divisible by 3.

Case 2: Let p=6k+5p = 6k + 5.

p2=(6k+5)2=36k2+60k+25p^2 = (6k + 5)^2 = 36k^2 + 60k + 25 p2+2=36k2+60k+27=3(12k2+20k+9)p^2 + 2 = 36k^2 + 60k + 27 = 3(12k^2 + 20k + 9)

which is divisible by 3. In both cases, p2+2p^2 + 2 is divisible by 3. ∴ Hence proved.

Exercise: EXERCISE A1.3 | Q: 3 | (Page 11)
For More Understanding

Explanation

The question asks to prove divisibility by 3 for p2+2p^2 + 2 where p≥5p \geq 5 is prime. The textbook context provides Example 11 which establishes that every prime greater than 3 must be of form 6k+16k + 1 or 6k+56k + 5 (since forms 6k6k, 6k+26k + 2, 6k+36k + 3, 6k+46k + 4 are divisible by 2 or 3 and hence not prime).

Using these two forms and expanding (6k+1)2+2(6k + 1)^2 + 2 and (6k+5)2+2(6k + 5)^2 + 2, we show both expressions have 3 as a factor.

Solution Steps

  1. Step 1: Since p≥5p \geq 5 is prime, p>3p > 3. By Example 11, p=6k+1p = 6k + 1 or p=6k+5p = 6k + 5 for some integer kk.

  2. Step 2: If p=6k+1p = 6k + 1, then

    p2+2=36k2+12k+3=3(12k2+4k+1)p^2 + 2 = 36k^2 + 12k + 3 = 3(12k^2 + 4k + 1)

    divisible by 3.

  3. Step 3: If p=6k+5p = 6k + 5, then

    p2+2=36k2+60k+27=3(12k2+20k+9)p^2 + 2 = 36k^2 + 60k + 27 = 3(12k^2 + 20k + 9)

    divisible by 3.

  4. Step 4: In both cases, p2+2p^2 + 2 is divisible by 3. ∴ Hence proved.

PreviousNext