You are missing the whole point. Its proof by induction.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?
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?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.
If you know that \(8\left|\left[4K^2+4K\right]\right.\) can you use that to prove \(8|\left[4K^2+12K+8\right]~?\)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!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?