If $\gcd(a,30)=1$, show that $60$ divides $a^4+59$

Question

anonymous · Answer

60 = 2^2 . 3 . 5

anonymous · Answer

$\large a$ is not even, so

$\large  a \equiv 1 \pmod 2 \implies a^2 \equiv 1 \pmod 4 \implies a^4 \equiv 1 \pmod 4$

ikram002p · Answer

Hint :-
a is odd thus

a=1 mod 4 
or 
a=-1 mod 4
both cases gives 

a^4=1 mod 4

now 
a^4=1 mod 4 (Fermat )
a^2=1 mod 3 
a^4=1 mod 3

anonymous · Answer

got it, thanks!

ikram002p · Answer

not welcome

ikram002p · Answer

i made a typo with next step
a^4=1 mod 5

anonymous · Answer

How is below argument without using odd/even thingy :

$\large  a = 1 \pmod 2\ \implies a^2 = 1 \pmod 2 \ \implies a^2-1 \equiv 0 \pmod 2 \ \implies (a+1)(a-1)\equiv 0 \pmod 2$

ikram002p · Answer

a is odd , so a-1 and a+1 both are even

anonymous · Answer

Exactly! so both are divisible by 2 and hence (a+1)(a-1) = 0 mod 2^2

ikram002p · Answer

yep.