Question

for every positive integer a , find a composite number n such that n|a^n-a.

Answer 1

Step-by-step explanation:

If n is even, n

4

+4

n

is divisible by 4

∴ It is composite number

If n is odd, suppose n=2p+1, where p is a positive integer

Then n

4

+4

n

=n

4

+4.4

2p

=n

4

+4(2

p

)

4

which is of the form n

4

+4b

4

, where b is a positive integer (=2

p

)

n

4

+4b

4

=(n

4

+4b

2

+4b

4

)−4b

2

=(n

2

−2b

2

)

2

−(2b)

2

=(n

2

+2b+2b

2

)(n

2

−2b+2b

2

)

We find that n

4

+4b

4

is a composite number consequently n

4

+4

n

is composite when n is odd.

Hence n

4

+4

n

is composite for all integer values of n> 1.

Answer 2

Given:$\text { If a is composite we may set } n=a \text { then } n \mid a^{n}+a \text {. if } a=1 \text { we may take for example } n=4$$\text { because } 1^{4}-1 \text { is divisible by } 4 \text {. When a is a prime greater than } 2 \text { we can take } n=2 a \text { because in }$

$\text { this case a is odd and even number } a^{2 n}-a \text { is divisible by } \mathrm{n} \text { and } 2 \text { and consequently by } 2 a=n \text {. }$$\text { It remains only the case where } a=2 . \text { In this case we may let } n=341=11 \times 31 \text {, which gives }$$341 \mid 2^{341}-2 \text { because: }$

$2^{10}-1 \equiv 0 \bmod 11$

$\Rightarrow 2^{340}-1 \equiv 0 \bmod 11$

$\text { which means } 2^{341}-2 \equiv 0 \bmod 11$

$\text { also: }$

$2^{5}-1 \equiv 2^{340}-1 \equiv 0 \bmod 31$

$\text { therefore: }$

$11 \times 31=341 \mid 2^{341}-2$