------------------------------------------------------------------------------------------------------------------

Solution to the third problem:

a^{7}-1

= a(a^{6}-1)

= a(a^{3}+1)(a^{3}-1)

= a(a+1)(a-1)(a^{2}+a+1)(a^{2}-a+1)

Clearly a(a+1) is divisible by 2 and a(a+1)(a-1) is divisible by 3.

Now the hard part is to show 7|(a^{7}-1). When divided by 7 the possible remainders are 1,2,3,...,6.

If a≡1(mod 7) then a^{3}≡1(mod 7).

If a≡2(mod 7) then a^{3}≡8(mod 7)≡1(mod 7)

If a≡3(mod 7) then a^{3}≡27(mod 7)≡-1(mod 7) i.e. a^{3}+1≡0(mod 7)

If a≡4(mod 7) then a^{3}≡64(mod 7)≡1(mod 7)

If a≡5(mod 7) then a^{3}≡125(mod 7)≡-1(mod 7) i.e. a^{3}+1≡0(mod 7)

If a≡6(mod 7) then a≡6(mod 7)≡-1(mod 7) i.e. a+1≡0(mod 7)

Therefore, the expression (a^{7}-1) is always divisible by 3 primes 2 , 3 and 7. So its divisible by 2x3x7=42 for all integers a.

--------------------------------------------------------------------------------------------------------------------------------

Solution to the fourth problem:

p and q are primes , p and q are distinct and a^{p }≡ a (mod q) and a^{q }≡ a (mod p).

By Fermat's Little Theorem we know a^{p }≡ a (mod p) and a^{q }≡ a (mod q).

Therefore, a^{p }≡ a (mod q) => (a^{p})^{q }≡ a^{q}(mod q) => a^{pq }≡ a^{q}(mod q) ≡ a (mod q)

Similarly a^{q }≡ a (mod p) => a^{pq }≡ a^{p}(mod q) ≡ a (mod q)

i.e ( a^{pq }- a) is divisible by both primes p and q where p and q are distinct. This implies that ( a^{pq }- a) is divisible by (pq) also.

So ( a^{pq }- a) ≡ 0 (mod pq) => a^{pq}≡ a (mod pq).

--------------------------------------------------------------------------------------------------------------------------------

Solution to the 5th problem:

p and q are primes , p and q are distinct.

Now

a^{p+q}- a^{p+1}- a^{q+1}+ a^{2}

= (a^{p}- a )(a^{q}- a)

= pK_{1}. qK_{2}for some integers K_{1}and K_{2}[By Fermat's Little Theorem]

= (pq).K where K=K_{1}.K_{2 }Therefore pq | (a^{p+q}- a^{p+1}- a^{q+1}+ a^{2})

--------------------------------------------------------------------------------------------------------------------------------