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

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

Answered by EDWIN88 last updated on 21/Feb/21

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

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