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Question Number 133320 by bramlexs22 last updated on 21/Feb/21

Find all n for which n^2 +2n+4 is divisible by 7

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{4}\: \\ $$$$\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\: \\ $$

Answered by EDWIN88 last updated on 21/Feb/21

let n = 7k+r then n^2 +2n+4 = (7k+r)^2 +2(7k+r)+4 = r^2 +2r+4 (mod 7) cheking for r = 0,1,2,3,4,5,6 we see that r = 1 or r = 4 Hence n is of the form 7k+1 or 7k+4 i.e n ≡ 1 or 4 (mod 7 ) test n=8 ⇒n^2 +2n+4 = 64+16+4=84 is divisible by 7 test n=11⇒n^2 +2n+4=121+22+4=147 is divisible by 7

$$\mathrm{let}\:\mathrm{n}\:=\:\mathrm{7k}+\mathrm{r}\:\mathrm{then}\:\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{4}\:=\:\left(\mathrm{7k}+\mathrm{r}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{7k}+\mathrm{r}\right)+\mathrm{4} \\ $$$$=\:\mathrm{r}^{\mathrm{2}} +\mathrm{2r}+\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{cheking}\:\mathrm{for}\:\mathrm{r}\:=\:\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\: \\ $$$$\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{r}\:=\:\mathrm{1}\:\mathrm{or}\:\mathrm{r}\:=\:\mathrm{4} \\ $$$$\mathrm{Hence}\:\mathrm{n}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{7k}+\mathrm{1}\:\mathrm{or}\:\mathrm{7k}+\mathrm{4}\: \\ $$$$\mathrm{i}.\mathrm{e}\:\mathrm{n}\:\equiv\:\mathrm{1}\:\mathrm{or}\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\:\right) \\ $$$$\mathrm{test}\:\mathrm{n}=\mathrm{8}\:\Rightarrow\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{4}\:=\:\mathrm{64}+\mathrm{16}+\mathrm{4}=\mathrm{84}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$$$\mathrm{test}\:\mathrm{n}=\mathrm{11}\Rightarrow\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{4}=\mathrm{121}+\mathrm{22}+\mathrm{4}=\mathrm{147}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$

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