Is 8|4(k^2 + k) true for k belonging to Z+?

[Mampac]

just got 8|4(k² + k).
well, the divisors of 8 are: 1, 2, 4, 8. Seems like I encompassed everything or do I miss something?

 

[anon_hedgepig]

The question is asking you if 8 divides the expression 4(k²+k) for all non-negative integers K.

I would recommend a proof by induction.

 

[Mampac]

the problem itself:
IF
n is odd and is a positive integer
THEN
n² ≡ 1 (mod 8)
________________________________
I started off with expressing n as 2k + 1 for some positive integer k, substituted in the congruent modulo and got stuck

 

[pka]

just got 8|4(k² + k).
well, the divisors of 8 are: 1, 2, 4, 8. Seems like I encompassed everything or do I miss something?

You are missing the whole point. Its proof by induction.
Is it true that \(8|4(1^2+1)\).
If we know that it is true that \(8|4(K^2+K)\) prove that \( 8|4([k+1]^2+[K+1])\) is true.

 

[Mampac]

You are missing the whole point. Its proof by induction.
Is it true that \(8|4(1^2+1)\).
If we know that it is true that \(8|4(K^2+K)\) prove that \( 8|4([k+1]^2+[K+1])\) is true.

something just occurred to me: if i factor k out i get 4k(k + 1). Now k(k + 1) is a product of two consecutive integers which is always even. Therefore, I can replace k(k + 1) with some 2t. Thus, I get 4*2t, that is 8|8t. Can I use this provided that my instructor proved that the product of two consecutive integers is always even earlier and wouldn't ask a proof of that?

 

[anon_hedgepig]

I think it's ironic that you're asking strangers on the internet what the expectations of your instructor are.

My advice is to just do the proof as a lemma (should be quick), or ask the instructor.

 

[pka]

something just occurred to me: if i factor k out i get 4k(k + 1). Now k(k + 1) is a product of two consecutive integers which is always even. Therefore, I can replace k(k + 1) with some 2t. Thus, I get 4*2t, that is 8|8t. Can I use this provided that my instructor proved that the product of two consecutive integers is always even earlier and wouldn't ask a proof of that?

If you know that \(8\left|\left[4K^2+4K\right]\right.\) can you use that to prove \(8|\left[4K^2+12K+8\right]~?\)

 

[HallsofIvy]

something just occurred to me: if i factor k out i get 4k(k + 1). Now k(k + 1) is a product of two consecutive integers which is always even. Therefore, I can replace k(k + 1) with some 2t. Thus, I get 4*2t, that is 8|8t. Can I use this provided that my instructor proved that the product of two consecutive integers is always even earlier and wouldn't ask a proof of that?

Either use the fact that the product of two consecutive integers is even "because your instructor proved that earlier" or prove it your self not noting that one of any two consecutive integers must be even!

If you really wanted to use "proof by induction" then, of course, you would first say that if k= 1 then 4(k^2+ k)= 4(1+ 1)= 4(2)= 8 which is obviously divisible by 8. Now assume that for some k= n, 4(n^2+ n) is divisible by 8, so that 4(n^2+ n)= 8m for some m, and use that to show that it is also true for k= n+1. That is, 4((n+1)^2+ (n+1))= 4(n^2+ 2n+ 1+ n+ 1)= 4(n^2+ n)+ 4(2n+ 2)= 8m+ 8(n+1)= 8(m+ n+ 1).