Appendix 1: Proofs in Mathematics | A1.3

Question 3

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

[Hint: Use Example 11].

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Solution
Understand the Question
  • By Euclid's division lemma, any positive integer when divided by 33 leaves a remainder of 0,1,0, 1, or 22, giving forms 3k3k, 3k+13k + 1, or 3k+23k + 2.
  • Since pp is a prime number and p≥5p \ge 5, pp cannot be divisible by 33, which rules out the form 3k3k.
  • Therefore, pp can only be of the form 3k+13k + 1 or 3k+23k + 2. We substitute both cases into p2+2p^2 + 2 and show that the result is always a multiple of 33.

Step 1 · Determine Possible Forms of Prime pp

Any integer can be written in one of the forms 3k3k, 3k+13k + 1, or 3k+23k + 2 for some integer kk.

Given that pp is prime and p≥5p \ge 5, pp cannot be a multiple of 33 (p≠3kp \neq 3k).

Therefore, pp must be of the form: p=3k+1orp=3k+2p = 3k + 1 \quad \text{or} \quad p = 3k + 2

Step 2 · Evaluate Case 1: p=3k+1p = 3k + 1

Substitute p=3k+1p = 3k + 1 into p2+2p^2 + 2

p2+2=(3k+1)2+2=(9k2+6k+1)+2=9k2+6k+3=3(3k2+2k+1)\begin{aligned} p^2 + 2 &= (3k + 1)^2 + 2 \\ &= (9k^2 + 6k + 1) + 2 \\ &= 9k^2 + 6k + 3 \\ &= 3(3k^2 + 2k + 1) \end{aligned}

Since 3(3k2+2k+1)3(3k^2 + 2k + 1) is a multiple of 33, p2+2p^2 + 2 is divisible by 33.

Step 3 · Evaluate Case 2: p=3k+2p = 3k + 2

Substitute p=3k+2p = 3k + 2 into p2+2p^2 + 2

p2+2=(3k+2)2+2=(9k2+12k+4)+2=9k2+12k+6=3(3k2+4k+2)\begin{aligned} p^2 + 2 &= (3k + 2)^2 + 2 \\ &= (9k^2 + 12k + 4) + 2 \\ &= 9k^2 + 12k + 6 \\ &= 3(3k^2 + 4k + 2) \end{aligned}

Since 3(3k2+4k+2)3(3k^2 + 4k + 2) is a multiple of 33, p2+2p^2 + 2 is divisible by 33.

Answer

Hence, for any prime number p≥5p \ge 5, p2+2p^2 + 2 is divisible by 33.

Common Mistakes
  • Including the 3k3k Case: Considering p=3kp = 3k even though p≥5p \ge 5 is given to be prime. The only prime divisible by 33 is 33 itself.
  • Algebraic Expansion Errors: Missing the middle term when squaring binomials, such as expanding (3k+1)2(3k+1)^2 as 9k2+19k^2 + 1 instead of 9k2+6k+19k^2 + 6k + 1.

More questions in A1.3

Q1

Prove that the sum of two consecutive odd numbers is divisible by 4.

Q2

Take two consecutive odd numbers. Find the sum of their squares, and then add 6 to the result. Prove that the new number is always divisible by 8.

Q3

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

[Hint: Use Example 11].

Q4

Let xx and yy be rational numbers. Show that xyxy is a rational number.

Q5

If aa and bb are positive integers, then you know that a=bq+ra = bq + r, 0≤r<b0 \le r < b, where qq is a whole number. Prove that HCF(a,b)=HCF(b,r)\text{HCF}(a, b) = \text{HCF}(b, r).

[Hint: Let HCF(b,r)=h\text{HCF}(b, r) = h. So, b=k1hb = k_1 h and r=k2hr = k_2 h, where k1k_1 and k2k_2 are coprime.]

Q6

A line parallel to side BCBC of a triangle ABCABC, intersects ABAB and ACAC at DD and EE respectively.

Prove that ADDB=AEEC\dfrac{\text{AD}}{\text{DB}} = \dfrac{\text{AE}}{\text{EC}}.

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