Prove that $a^{4b+c} + a^{4b+d}\equiv 0[30]$

Question

Assume that $a, b,c,d$ are positive integers. I'm confused about this claim: $$a^{4b+c} + a^{4b+d}\equiv 0[30]$$. I think I will be consider this cases:

1. $a^{4b+c} + a^{4b+d}\equiv 0[2]$, true because, by disjonction cases, we have $a^{4b}(a^c+a^d) \equiv 0[2]$.
2. But I have no idea to do withis two cases $a^{4b+c} + a^{4b+d}\equiv 0[3]$ and $a^{4b+c} + a^{4b+d}\equiv 0[5]$.

Thank's in advance, and any help is welcome.

Answer 1

It's not necessarily true. Take for example $a=b=c=d=1$.

Answer 2

What? Of course it isn't true.

wolog Let $d = c +k$ so $a^{4b + c} + a^{4b + d}= a^{4b + c}(1 + a^k)$. There is utterly no reason that should be congruent to $0\mod 30$ and several counter examples you can come up with. $2^{4*1 + 1}(1 + 2^2)= 32*5 \equiv 10 \mod 30$, for instance if $a = 2; b=1; c = 1, d=3$.

Anyway... to do $\mod 5$. It's true if $a \equiv 0 \mod 5$ but otherwise $a^4 \equiv 1 \mod 5$ so $a^{4b + c} + a^{4b + d} \equiv a^c + a^d$ which obviously do not need to be equivalent to $0 \mod 5$.

Likewise for $\mod 3$. and $a \not \equiv 0 \mod 3$ then $a^{4b + c} + a^{4b + d} \equiv a^c + a^b$ which need not be equiv $0$.

I figure that and as $4 = 4\phi(2) = 2\phi(3)=\phi(5)$ that this exercise has been misstated and it's supposed to be something with some relation between $c$ and $d$, in which case this could be proven with CRT but.... I don't see what the correct stating of the problem would be.