**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.

*Related*

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.

Hmm it would … But then you have to check that term mod 7

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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How much marks are required in paper2 of b.maths???