# I.S.I. 2015 subjective solution (Problem 2)

Find all such Natural number n such that 7 divides $$5^n + 1$$

Discussion:

Teacher: This is simple case of modular arithmetic. Consider the set of all residues of $$5^n$$ modulo 7 and you will see a pattern

Student: Sure. $$5^1 = 5 , 5^2 = -3, 5^3 = -1 , 5^4 = 2 , 5^5 = 3 , 5^6 = 1$$ modulo 7. The residues repeat after that since $$5^{6k + m} = 5^{6k} \times 5^m = 5^m$$ modulo 7 where $$m \le 6$$ .

We want $$5^n = -1$$ mod 7 hence n must be an odd multiple of 3 or n = 6r + 3 for any nonnegative integer r.

## 5 Replies to “I.S.I. 2015 subjective solution (Problem 2)”

1. so ashani sir i could do the sum in following way that if i write 5 as (7-2)^n and expand it binomially then we have a term indepent of 7 and that term if i calculate so it would give me the answer possibly.

1. yes that term mod 7 will give us the remainder part which is not a multiple of 7 and in that part if we put value of n i think probably 2 i dont remember as in paper i have done and that would give the term multiple of 7 and so we will get its value mean the value of n
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