# Algebraic Arguments

• June 6th 2012, 12:23 PM
curiousone
Algebraic Arguments
Hi Guys,
I'm doing GCSE maths higher tier. And generally I'm quite good at algebra, subsitiution, factorising etc but I'd appreciate help with arguments.

Here is the question:
"Prove that the sum of squares of any two consecutive integers is an odd number".

It has to be done algebraically. I have the answers but I don't understand them, can you please describe how you get the answer, also
please can you provide general tips for answering these arguments mental approach, algebra etc. and the quicker the reply the better. I have a maths
exam and it's quite urgent, so all help will be appreciated.

Thanks,
• June 6th 2012, 12:32 PM
skeeter
Re: Algebraic Arguments
Quote:

"Prove that the sum of squares of any two consecutive integers is an odd number".
let $x$ = initial integer

$x+1$ = next consecutive integer

$x^2 + (x+1)^2 = x^2 + (x^2 + 2x + 1) = 2x^2 + 2x + 1 = 2(x^2 + x) + 1$

now, what can you say about the quantity $2(x^2+x)$ , where $x$ is any integer?

what happens when you add 1 to that quantity?
• June 6th 2012, 12:38 PM
emakarov
Re: Algebraic Arguments
Quote:

Originally Posted by curiousone
"Prove that the sum of squares of any two consecutive integers is an odd number".

This is similar to proving that the sum of two consecutive integers is an odd number. This is, of course, because one of these integers is odd and the other is even. The reason the original question is similar is that the square of an even number is even and the square of an odd number is odd.

Quote:

Originally Posted by curiousone
can you provide general tips for answering these arguments mental approach, algebra etc.

When you are asked about two consecutive integers, you consider n and n + 1 for some n. An even number has the form 2k for some integer k, while an odd number has the form 2k + 1 for some integer k. For example, here is a proof that the square of an odd number is odd. Suppose you are given an arbitrary odd number. Then there must exist an integer k such that the given number is 2k + 1. Therefore, its square is (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2) + 1 = 2k' + 1 for k' = 2k2 + 2. Since k' is an integer, 2k' + 1 is odd.
• June 6th 2012, 12:39 PM
curiousone
Re: Algebraic Arguments
Of what you said i understand x(squared)+(x{squared} +2x +1) from then on I'm lost.
• June 6th 2012, 12:43 PM
curiousone
Re: Algebraic Arguments
Thanks for the help on the mental and algebraic bit :)
• June 6th 2012, 01:52 PM
Deveno
Re: Algebraic Arguments
here's everything you need to know about even- and oddness:

even + even = even
even + odd = odd
odd + odd = even

(even)*(even) = even
(even)*(odd) = even
(odd)*(odd) = odd

if we have two consecutive integers, one is even, and one is odd. therefore, their squares are (even)*(even) = even and (odd)*(odd) = odd.

when we add their squares, we get: even + odd = odd.