> # arp -s 1.5.1.1 00:a0:c9:39:60:27 pub
> cannot intuit interface index and type for 1.5.1.1

I don't know whether it will work with a _proxy_ arp entry, but when I
was playing with arp entries and such, I found them to be
interface-specific, but this specificity was not reflected back to the
command-line by the provided tools.  (Note that the message implies
that arp is trying to "intuit" some stuff, but gives you no way to
override this in case you want something other than its guess, or in
case its intuiting fails.

You may find yourself needing to throw together a tool that gives you
command-line access to the specific interface you need.  (If even that
works, of course; it might not.)  I can send you code extracted from a
program I wrote that adds arp entries pointing out "unusual"
interfaces.  Unlike your case, the address I wanted to add an entry for
was one the box had, but arp's guess was wrong for my application.

One thing that might work is to add the address you want to the
interface you want, add the arp entry, and then delete the address:

# ifconfig fxp0 alias 1.5.1.1/32
# arp -s 1.5.1.1 ..:..:..:..:..:.. pub
# ifconfig fxp0 delete 1.5.1.1

Or perhaps 1.5.1.0/28 or something instead.

> is it not possible to proxyarp for a foreign address?

I think the theory is that if the address wouldn't normally fall on
that cable, nobody will arp for it on that cable.  Then "wouldn't
normally fall on that cable" is equated to "we don't have an interface
address implying it falls on that cable", and the problem you see
arises.

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