Appendix 1: Proofs in Mathematics | A1.3

Question 2

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.

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Solution
Understand the Question
  • Any odd number can be represented algebraically in the form 2n−12n - 1 (or 2n+12n + 1), where nn is an integer.
  • Two consecutive odd numbers differ by 22, so we can represent them as 2n−12n - 1 and 2n+12n + 1.
  • We compute the sum of their squares, (2n−1)2+(2n+1)2(2n - 1)^2 + (2n + 1)^2, and add 66 to the resulting expression.
  • If the simplified result can be written as 8×(integer)8 \times (\text{integer}), it proves the number is always divisible by 88.

Step 1 · Represent the Consecutive Odd Numbers

Let the two consecutive odd numbers be (2n−1)(2n - 1) and (2n+1)(2n + 1), where nn is an integer.Diagram 1

Step 2 · Find the Sum of Their Squares

Squaring each number

(2n−1)2=(2n)2−2(2n)(1)+12=4n2−4n+1\begin{aligned} (2n - 1)^2 &= (2n)^2 - 2(2n)(1) + 1^2 \\ &= 4n^2 - 4n + 1 \end{aligned} (2n+1)2=(2n)2+2(2n)(1)+12=4n2+4n+1\begin{aligned} (2n + 1)^2 &= (2n)^2 + 2(2n)(1) + 1^2 \\ &= 4n^2 + 4n + 1 \end{aligned}

Adding the two squares

(4n2−4n+1)+(4n2+4n+1)=4n2+4n2−4n+4n+1+1=8n2+2\begin{aligned} (4n^2 - 4n + 1) + (4n^2 + 4n + 1) &= 4n^2 + 4n^2 - 4n + 4n + 1 + 1 \\ &= 8n^2 + 2 \end{aligned}

Step 3 · Add 6 to the Result

Adding 66 to the sum of squares

(8n2+2)+6=8n2+8\begin{aligned} (8n^2 + 2) + 6 &= 8n^2 + 8 \end{aligned}

Step 4 · Prove Divisibility by 8

Factoring out 88

8n2+8=8(n2+1)\begin{aligned} 8n^2 + 8 &= 8(n^2 + 1) \end{aligned}

Since nn is an integer, (n2+1)(n^2 + 1) is also an integer.

Therefore, 8(n2+1)8(n^2 + 1) is a multiple of 88 and is always divisible by 88.

Answer

Hence proved that the new number 8(n2+1)8(n^2 + 1) is always divisible by 88.

Common Mistakes
  • Incorrect Algebraic Setup: Assuming consecutive odd numbers are nn and n+2n + 2 without defining nn as an odd integer, or using 2n+12n + 1 and 2n+22n + 2 (which would be an odd and an even number).
  • Expansion Errors: Forgetting the cross-term when expanding squares, such as writing (2n−1)2(2n - 1)^2 as 4n2−14n^2 - 1 instead of 4n2−4n+14n^2 - 4n + 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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