I at least don't know of any way to do it without it.

I think the example of cryptography is a little bit more complicated. Mathematicians working in cryptography are usually very careful to state as a hypothesis that certain problems are difficult (or just that there are difficult problems) much in the same way that certain results are conditional on the Riemann hypothesis. On the other hand physicists working in quantum cryptography seem to use "prove" in the sense of physics. Hence one sees statements to the effect that quantum cryptography is proved to be secure whereas RSA is not.

The universal coefficient formula for cohomology says that $H^i(X,M)$ is isomorphic to $\mathrm{Hom}(H_i(X,\mathbb Z),M)\bigoplus\mathrm{Ext}^1(H_{i-1}(X,\mathbb Z)$. For $M=\mathbb Z$ that means that the torsion free part of $H^i(X,\mathbb Z)$ is dual (in the sense of $\mathrm{Hom}(-,\mathbb Z)$ to the torsion free part of $H_i(X,\mathbb Z)$ whereas the torsion of $H^i(X,\mathbb Z)$ is dual (in the sense of $\mathrm{Hom}(-,\mathbb Q/\mathbb Z)$ to the torsion part of $H_{i-1}(X,\mathbb Z)$.

(cont'd) If you do it for $\mathbb Q/\mathbb Z$-coefficients it doesn't but you lose the precise integral information (as you cannot canonically recover a finitely generated $\mathbb Z$-module from its $\mathbb Q/\mathbb Z$-dual.

It is easier to understand what happens to the torsion if you write Poincaré duality as an isomorphism between homology and cohomology; $H^i(M;\mathbb Z)=H_{n-i}(M,\partial M;\mathbb Z)$. To get a formula involving only cohomology you then combine that with the universal coefficient formula. If you do that for $\mathbb Z$-coefficients, the torsion moves around a little bit as it comes from an $\mathrm{Ext}^1$. (cont'd)

Indeed, for $p>2$ the reduction is injective and for $p=2$ the kernel consists of elements of order $2$ for a general $\mathrm{GL}_n(\mathbb Z)$. Hence, in the case of $\mathrm{SL}_2(\mathbb Z)$ we get that the kernel has at most order $2$ and order of a finite subgroup is a divisor of $2\cdot|\mathrm{SL}_2(\mathbb Z)/2|=12$. If we restrict ourselves to commutative subgroups we get a divisor of $4$ or $6$ (as the order of elements of $\mathrm{SL}_2(\mathbb Z)/2$ are $1$, $2$ or $3$) but that also follows from the fact that these are the only $d$ with $\varphi(d)\leq 2$.

@Martin: I did as you suggested (with some elaboration to further justify the move). I put this here so that your answer makes sense as I delete my comments here.