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I need help with getting on the right track for this one.

Since $a = b \; \text{mod} \; n$, we have that $a = kn + b$ for some $k \in \mathbb{Z}$. Likewise, we have $c = jn + d$ for some $j \in \mathbb{Z}$. Therefore, we must show that $n | a^c - b^d$. However, we can substitute $(kn+b)$ for $a$ and $jn+d$ for $c$ to get $(kn+b)^{jn+d} - b^d$. By the binomial theorem, we have $\sum_{i = 0}^{jn+d} \binom{jn+d}{i}(kn)^{i}d^{jn+d-i} - b^d.$ It is at this point I get stuck though. Am I on the wright track, or is there a much easier way?

Edit: A slight modification, the homework did say $a,b,c,d$ were all positive integers.

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    What makes you think it's true? Have you tested it? – 2017-02-20
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    It's an assignment from the homework, so I assumed it to be true. I guess I shouldn't trust that though – 2017-02-20
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    Perhaps it was asked as a question (i.e., "if ... must it be true that .. ?"), Or maybe it was phrased as "prove or disprove ...". In any case, the lesson here is if you're asked to prove something that's not obviously true, at least do some spot checks. – 2017-02-20
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    Double checking it, the way I phrased it is correct. Maybe it was a typo or an oversight. – 2017-02-20
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    If $a=b$ and 4d=c+n4 and 4\gcd (a,n)=14 then 4a^c\equiv b^d \pmod n \iff 0\equiv a^c(a^n-1)\pmod n \iff a^n-1\equiv 0\pmod n.$ When $a=2$ and $n=3$ we have $a^n-1=7 \not \equiv 0 \pmod 3$. – 2017-02-20
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    $3^2 \equiv 2 \mod 7$, $3^{2 + 7}=3^9 = (3^3)^3 = 27^3 \equiv -1^3 \equiv -1 \equiv 6 \mod 7$. So for $3 \equiv 3 \mod 7$ and $2 \equiv 9 \mod 7$ it does not appear to be true. – 2017-02-20
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    Thing is... this implies $a^n \equiv a^0 = 1 \mod n$ which... if you are familiar with Eulers thereom isn't true in general. – 2017-02-20

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This is not true! Consider $a = b=2$ $c = 0$, $d = 7$, $n = 7$.

Then $a^c = b^c = 1 \pmod 7$, but $a^7 = b^7 = 2 \pmod 7$

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    Ah, thank you so much! – 2017-02-20
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    What if we modify it so that a,b,c,d are all positive integers? – 2017-02-20
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    No, $2^1 \neq 2^8 \pmod 7$ – 2017-02-20
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The thing is... if this were true this would imply $a^{n+b} = a^na^b \equiv a^b \mod n$ and if $b=0 \implies a^n \equiv 1 \mod n$ which... doesn't seem right. In fact Fermat's Little Theorem and Euler's Theorem say it is not true.

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    Didn't I say that FLT --- faster than light travel--- Fermat's Little Theorem just what I wro.... oh, googly-moogly! – 2017-02-20