Q3/017) If 30 and a are co primes then show that 60 divides a^2+ 59

30 and a are co primes so 2 , 3  5 none of these divide a

Now a^4-1 = (a+1)(a-1)(a^2 +1)

As a is odd so a+1 and a-1 are even and product is divisible by 4

As a is not divisible by 3 so (a+1) or (a-1) is divisible by 3

Further  a^4-1 mod 5 = (a+1)(a-1)(a^2 +1) mod 5

= (a+1)(a-1)(a^2-4) mod 5

= (a+1)(a-1)(a+2)(a-2) mod 5

Now (a-2), (a-1), a , (a+1), (a+2) being 5 consecutive numbers so one of them is divisible by 5 and a is not so one of the rest must be divisible

So a^4-1 is divisible by 5.

So   a^4-1 is divisible by 4 ,3 and 5 and hence product 60

So a^4-1 + 60 or a^4+ 59 is divisible by 60