Answered: "Let D=a2+b2+c2 where a and b are successive..."

ebnspmia0jj

Answered question

2022-02-03

Let

D = a^{2} + b^{2} + c^{2}

where a and b are successive positive integers and

c = a b

. How will you show that

\sqrt{D}

is an odd positive integer?

ashuhra6e

Beginner2022-02-04Added 15 answers

D = {(a^{2} + a + 1)}^{2}

which is the square of an odd integer.
Explanation:
Given a, we have:

b = a + 1

c = a b = a (a + 1)

D = a^{2} + {(a + 1)}^{2} + {(a (a + 1))}^{2}

= a^{2} + (a^{2} + 2 a + 1) + a^{2} (a^{2} + 2 a + 1)

= a^{4} + 2 a^{3} + 3 a^{2} + 2 a + 1

= {(a^{2} + a + 1)}^{2}

If a is odd then so is

a^{2}

and hence

a^{2} + a + 1

is odd.
If a is odd then so is

a^{2}

and hence

a^{2} + a + 1

is odd.

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