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According to a source online, the method to prove a statement $P(m,n)$ to be true the base case is to show that both $$P(1,n) \ \ P(m,1) $$ hold for all $m,n \in \mathbb N$. The inductive step is that if both $$P(m+1,n) \ \ P(m,n+1)$$ hare assumed to hold then $P(m+1,n+1)$ holds for all $m,n \in \mathbb N$.

So if my $P(1,n) \ \ P(m,1)$ imply that only odd integer solutions exist and if $P(m+1,n) \ \ P(m,n+1)$ also implies that only odd integer solutions exist and that since they are odd intger solutions then $P(m+1,n+1)$ holds. But I'm not sure what would imply that only odd integer solutions exist for $P(m+1,n+1)$

argamon
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    This is not clear. Where does the parity of the solutions enter into it? – lulu Oct 11 '20 at 17:27
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    For whatever it's worth: I would not say that this was the $\textit {only}$ route to proving $P(n,m)$ inductively. Often it works to simply show $P(1,1)$ and then show that $P(n,k)$ implies $P(n+1,k)$ and $P(n, k+1)$ or something like that. – lulu Oct 11 '20 at 17:33
  • @lulu that is helpful I did not think of that – argamon Oct 11 '20 at 17:38

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