Let be a symmetric matrix of order $(2n+1)×(2n+1)$ which has on each row and each column every integer from $1$ to $2n+1$. What is the trace of the matrix?
So, my guess is that the trace has every integer from $1$ to $2n+1$, so it is $(2n+1)(n+1)$ and I am almost sure that I am right because I have studied some particular cases with 3 and 5, and I was even able to find some proofs for that. I also have tried to generalize, but I wasn't even able to do it for $7$, so I am clearly not ready to solve the problem for $2n+1$. Please, any hint will be appreciated because I have to solve this problem by tomorrow.
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Gary
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2Hint: Note that $a_{ij}=a_{ji}$. Therefore, if you look at the entries equal to $1$, there are an even number outside the diagonal. This shows that in the diagonal there is an odd number of entries equal to $1$. Can you generalize this idea? – Nicolás Vilches Jan 15 '22 at 22:58
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1These matrices are called "Latin squares". Important hint; https://math.stackexchange.com/q/2623403 – Jean Marie Jan 15 '22 at 23:05
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@NicolásVilches That is a lovely hint – Mark Bennet Jan 15 '22 at 23:06
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1I have taken the liberty to change your title in order for it to reflect in a bettre way the content of your question, and hopefuly to attract more people on it. Do you agree ? – Jean Marie Jan 15 '22 at 23:10
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1@Jean Marie Yes, I agree. And thanks for your help. – Lucas McAllister Jan 15 '22 at 23:14
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@JeanMarie I think that "symmetric" is essential here. In order 3 a Latin square will have three identical digits along one of the diagonals. Only the "symmetric" condition means this cannot be the leading diagonal. I am about to add the word "symmetric" to the title. – Mark Bennet Jan 15 '22 at 23:26
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@Mark Bennet You are perfectly right. Please do it. – Jean Marie Jan 15 '22 at 23:27