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My Question Reads:

If $a, b$ are integers such that $a \equiv b \pmod p$ for every positive prime $p$, prove that $a = b$.

I started by stating $a, b \in \mathbb Z$.

From there I have said without loss of generality, $a \geq b$.

Suppose $1 \leq a - b \in \mathbb Z$.

From here I expressed $a - b$ as a prime factorization.

$a - b = p_1 \times p_2 \times \ldots \times p_n$

From here I said to consider a prime $p$ such that $p$ does not equal $p_k$ for all $1$ less than or equal to $k$ less than or equal to $n$. From here I am a bit lost as to how to continue.

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    **Hint** $ $ "prime" is a red herring. All we need is for it to hold for arbitrarily large naturals $p.\, $ What can we deduce from $\,p\mid a\!-\!b\,$ by choosing $\,p> |a\!-\!b|\,?\ \ $ – 2017-01-25
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    @BillDubuque how do you know p>|a-b| ? – 2017-01-25
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    By our (new) hypothesis, for any integer $n$ there is some $\,p> n\,$ such that $\,p\mid a\!-\!b.\,$ Let $\,n=|a-b|.\ \ $ – 2017-01-25
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    It boils down to the fact that zero is the only natural with *unbounded* (prime) factors. But we don't need "prime" to prove that - which is why I enclosed it in parentheses. – 2017-01-25
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    @BillDubuque is there any way to do this taking the sort of approach I did by writing out a prime factorization of a-b? I understand it better this way because I am only just learning this – 2017-01-25
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    Sure, just choose $p$ to be larger than any prime dividing $|a-b|.\,$ But it is serious overkill to use prime factorizations to deduce the simple fact that a larger natural cannot divide a smaller natural (unless zero). – 2017-01-25
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    @BillDubuque yes, I think it is the longer way, but this is all new material to me so I might understand it better – 2017-01-25
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    @BillDubuque I see that would work but I am not sure how to set it up to show this exactly – 2017-01-25
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    I strongly recommend against that The simpler way is a *very* common proof that works *far* more generally than does using prime factorizations. If you don't learn this simple ubiquitous argument then you will be stuck on other problems where the method using primes does not work. – 2017-01-25
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    @BillDubuque well for now this is how I am learning it, but once I know more I will definitely move away from this method – 2017-01-25
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    I'm assuming you're new here, Sam. For now, I'm willing to improve your formatting. But I expect you to learn to do it yourself. – 2017-01-25

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Suppose $a \neq b$. Then $b-a \neq 0$, WLOG $b > a$, so there is some prime $p > b-a > 0$ (primes are infinite), but then $a \not \equiv b \mod p$, obviously. This is a contradiction, hence we are done i.e. $a=b$.