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

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#6

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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#7

(Original post by

Induction is the real man’s approach!

**RDKGames**)Induction is the real man’s approach!

(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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#8

(Original post by

So many full solutions in breach of the rules, and yet no-one using difference of 2 squares to answer this in 1 line...

**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...

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#9

(Original post by

(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.

**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.

For we have:

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#10

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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#11

(Original post by

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?

**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?

How this is done in practice varies. It didn't

**need**to be done in your way. You essentially did 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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#12

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.

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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#13

(Original post by

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 is div by 4 while knowing that holds true, and assuming that holds true.

How this is done in practice varies. It didn't

**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 is div by 4 while knowing that holds true, and assuming that holds true.

How this is done in practice varies. It didn't

**need**to be done in your way. You essentially did 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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#14

(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

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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#15

(Original post by

(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

**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

(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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