Prove that (n+5)^2-(n+3)^2 is a multiple of 4 for all positive integer value of n

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jjasn
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Please could you put your working out thankssss
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username3648614
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Expand, simplify and then prove.
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Texxers
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Expand it, simplify it and then take 4 out as a factor.
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RDKGames
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Induction is the real man’s approach!
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Notnek
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Full solutions are a no no.
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DFranklin
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So many full solutions in breach of the rules, and yet no-one using difference of 2 squares to answer this in 1 line...

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Amullai
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(Original post by RDKGames)
Induction is the real man’s approach!
n=1
(1+5)^2-(1+3)^2 = 36-16 = 20
20/4 = 5
So the result is true for n=1

(k+5)^2-(k+3)^2 = 4 times an integer
Can't really do proof by induction from here on(unless I'm mistaken) so I'm going to work out the difference and prove that that is also a multiple of 4.

(k+1+5)^2 -(k+5)^2 = 2x1x5+2x1xk+1x1
(k+1+3)^2 -(+3)^2 = 2x1x3 +2x1xk +1x1
10-6 =4
2k-2k=0
1-1=0
So the result if the result is true for k, then the result is true for k+1.
Since the result is true for n=1, the result is true for all positive integers greater than or equal to 1 by induction(or at least by this method).
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username3249896
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(Original post by DFranklin)
So many full solutions in breach of the rules, and yet no-one using difference of 2 squares to answer this in 1 line...
One of the solutions did use the difference of 2 squares.
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RDKGames
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(Original post by Amullai)
(k+5)^2-(k+3)^2 = 4 times an integer
Can't really do proof by induction from here on(unless I'm mistaken) so I'm going to work out the difference and prove that that is also a multiple of 4.
Sure you can!

For n=k+1 we have:
(k+1+5)^2-(k+1+3)^2=[(k+5)^2+2(k+5)+1]-[(k+3)^2+2(k+3)+1]

\displaystyle =\underbrace{(k+5)^2-(k+3)^2}_{\mathrm{div. \ by \ 4 \ from \ assumption}}+4
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Amullai
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(Original post by RDKGames)
Sure you can!

For n=k+1 we have:
(k+1+5)^2-(k+1+3)^2=[(k+5)^2+2(k+5)+1]-[(k+3)^2+2(k+3)+1]

\displaystyle =\underbrace{(k+5)^2-(k+3)^2}_{\mathrm{div. \ by \ 4 \ from \ assumption}}+4

Isn't that what I did? I thought that induction had to be by adding the difference and then factorising out things until you get to a point where you make each factor for k+1? Or is it the same thing just done in a different way?
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RDKGames
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(Original post by Amullai)
Isn't that what I did? I thought that induction had to be by adding the difference and then factorising out things until you get to a point where you make each factor for k+1? Or is it the same thing just done in a different way?
Sort of. Induction is the principle of proving that a statement holds true for the 'next' item along, using the assumption that it holds true on your 'current' item. So it's proving that f(k+1) is div by 4 while knowing that f(1) holds true, and assuming that f(k) holds true.

How this is done in practice varies. It didn't need to be done in your way. You essentially did f(k+1)-f(k) and shown that this is a multiple of 4, in a weird way that took me two or three reads to figure out.
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math42
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Difference of two squares is cool and all, but the only reasonable method is mods.
If k is even, k^2 = 0 mod 4.
If k is odd, k^2 = 1 mod 4.
n + 5 and n + 3 have the same parity.
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Amullai
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(Original post by RDKGames)
Sort of. Induction is the principle of proving that a statement holds true for the 'next' item along, using the assumption that it holds true on your 'current' item. So it's proving that f(k+1) is div by 4 while knowing that f(1) holds true, and assuming that f(k) holds true.

How this is done in practice varies. It didn't need to be done in your way. You essentially did f(k+1)-f(k) and shown that this is a multiple of 4, in a weird way that took me two or three reads to figure out.
Ok thanks.
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lucia_gester
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(n+5)(n+5)
n²+5n+5n+25-((n+3)(n+3))
n²+3n+3n+9
n²+10n+25-(n²-6n-9)
4n+16n
/4
4(n+4)
divsble by 4 hence i is a multiple
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DFranklin
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(Original post by lucia_gester)
(n+5)(n+5)
n²+5n+5n+25-((n+3)(n+3))
n²+3n+3n+9
n²+10n+25-(n²-6n-9)
4n+16n
/4
4(n+4)
divsble by 4 hence i is a multiple
Difference of 2 squares: (n+5)^2-(n+3)^2 = ((n+5)-(n+3)) ((n+5)+(n+3)) = 2(2n+8) = 4(n+4).

(But please don't reopen 2 year old threads. Also posting full solutions is generally against the rules - this thread was a bit of an exception).
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