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Apr 2, 2019 at 2:53 comment added Timothy Chow @KConrad : Provided one takes some care in formulating the pigeonhole principle, one can prove it (by induction) in Heyting arithmetic, and hence in a certain precise sense it does not rely on proof by contradiction. See for example "Ramsey's theorem and the pigeonhole principle in intuitionistic mathematics" by Veldman and Bezem.
Oct 13, 2018 at 4:49 comment added KConrad One of the conditions set by the OP was that the argument does not use contradiction, and I think anything involving the pigeonhole principle is in some sense relying on a proof by contradiction: if $f : A \rightarrow B$ and $|A| > |B|$ then $f$ can't be injective, because if $f$ were injective then $|A| \leq |B|$ and we have a contradiction. I was considering posting an argument with the pigeonhole principle earlier but chose not to due to the OP's desire not to see proofs that somehow involve contradiction.
Oct 13, 2018 at 2:23 history answered Timothy Chow CC BY-SA 4.0