Question

Prove that the sum of the squares of two consecutive integers is odd

Answer 1

Answer: Every even number can be expressed in the form (2n) where n is an integer. So we have an even number, 2n, and an odd number, 2m+r. Adding them together we get 2n + 2m + r, which can be expressed as 2 (n+m) + r.

Answer 2

Answer:

Step-by-step explanation:

Start by breaking up the question into parts:'always an odd number' . An even number is any number divisible 2 e.g. 2n, an odd number is always 1 higher or lower than an even number e.g. 2n+1 or something similar. Getting an answer at the end of working of this form will prove the theory'any two consecutive integers' means two numbers that are next to each other on the number line.To prove this, we can do this in terms of n i.e. n and n+1'the squares' square the two terms we have just written down to give n2 and (n+1)2The 'sum of the squares' tells us to add these two parts together. We would then hope to work through this and get something similar to the idea of 2n+1 or something similarn2+(n+1)2=2n2+2n+1=2(n2+n) +1. The first part of this is even because it is divisible by 2. By adding 1 it must become odd.