Let N be an integer not divisible by 3. Prove N^2 = 3a + 1, where a is an integer

For N to be not divisible by 3, N can either be of the form 3k + 1 (1,4...) or 3k + 2 (2,5...), where k is an integer.
The proof can then be done by checking both 3k + 1 and 3k + 2 when N is squared, to see if they can be rearranged into the form 3a + 1.
N = 3k + 1, so N2 = (3k + 1)2 = 9k2 + 6k + 1
This can then be rearranged to prove 3a + 1. Note that a can be made of any polynomial of k with integer powers, as k is an integer so its polynomial with integer powers will also be an integer for any value of k.
9k2 + 6k + 1 = 3(3k2 +2k) + 1, so true for N = 3k + 1
The same method can then be used to prove for N = 3k + 2
N= 3k + 2, N2 = 9k2 + 12k + 4 = 3(3k2 + 4k + 1) + 1

Answered by Maths tutor

24046 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Solve the equation sin2x = tanx for 0° ≤ x ≤ 360°


Simplify the following expression to a fraction in its simplest form: [(4x^2 + 6x)/(2x^2 - x -6)] - [(12)/(x^2 - x - 2)]


How can I recognise when to use a particular method for finding an integral?


Express the fraction (p+q)/(p-q) in the form m+n√2, where p=3-2√2 and q=2-√2.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences