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Jul 9 at 13:01 comment added Ian Wanless Indeed, I goofed (rushed my reply). Have now edited the original post, with my next attempt at counterexamples.
Jul 9 at 12:59 history edited Ian Wanless CC BY-SA 4.0
added 124 characters in body
Jul 4 at 16:03 history edited Ian Wanless CC BY-SA 4.0
deleted 820 characters in body
Jul 4 at 14:47 comment added Richard Stanley @Ian Wanless: your $F$ looks like a Latin square to me. This is the case $b=1$. Every $a\times a$ Latin square certainly contains an $a\times a$ Latin square, namely, itself.
Jul 4 at 13:23 history answered Ian Wanless CC BY-SA 4.0