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I am trying to prove that for positive integers s and t and u is a vector, if diag $(suu^T)\equiv$ diag$(tuu^T)\pmod n$ then $suu^T\equiv tuu^T\pmod n$. So what I was trying to prove is

If for all i, $su_i^2\equiv tu_i^2\pmod n$, then $su_iu_j\equiv tu_iu_j\pmod n$.

G.May
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2 Answers2

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Let $\,a = s\!-\!t.\,$ Then $\,n\mid au^2,av^2\,\Rightarrow\, n\mid a(u^2,v^2) = a(u,v)^2 = a(u^2,uv,v^2)\,\Rightarrow\,n\mid auv.\ $ QED

Or use $\ n^2\mid (auv)^2\,\Rightarrow\, n\mid auv,\,$ provable by unique prime factorization, or the Rational Root Test, i.e. $\, x = (auv/n)\in\Bbb Q\,$ is a root of $\,x^2 = k\in\Bbb Z\,$ so $\,x\in \Bbb Z$.

Bill Dubuque
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  • Sorry. It's just that I am not familiar with the notation you used. What do you mean by $n | a(u^2,v^2)= a(u,v)^2$. Thank you for your help. – G.May Feb 26 '20 at 08:26
  • I think the notation $(u,v)$ means $\gcd(u,v)$. As to the equalities $(u^2,v^2)=(u,v)^2=(u^2,uv,v^2)$ they are some elementary exercises. :p – awllower Feb 26 '20 at 10:11
  • @G.May Yes $(m,n)$ denotes $\gcd(m,n).\ $ $,$ We used basic gcd laws, e.g. see here. $\ x\mid y$ means $x$ divides $y$. This is all standard number theory notation. – Bill Dubuque Feb 26 '20 at 15:30
  • Oh. Okay. I understand now. Thank you. – G.May Feb 27 '20 at 00:48
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If for some $i$, $u_i$ is co-prime to $n$, then there is an integer $u_i^{-1}$ such that $u_i^{-1}u_i\equiv1\pmod n$. This implies that $s\equiv t\pmod n$, and the statement obviously holds in this case. Similarly, if $s\equiv t\pmod n$, the we are done.

So suppose $\gcd(u_i,n)\ne1,\,\forall i$ and $s\not\equiv t\pmod n$. Then we are given that $(s-t)u_i^2\equiv0\pmod n$. This means $\frac{n}{\gcd(n,s-t)}\mid u_i^2$. Denote $\frac n{s-t}=m$ and write its prime factorization as $$m=\prod_kp_k^{n_k}.$$ Then we deduce that $$\prod_kp_k^{\lceil\frac{n_k}2\rceil}\mid u_i,\,\forall i.$$ Consequently, $$m=\prod_kp_k^{n_k}\mid u_iu_j.$$ Therefore $n=\gcd(n,s-t)\cdot m\ \mid(s-t)\cdot u_iu_j$, i.e. $su_iu_j\equiv tu_iu_j\pmod n$.

Hope this helps.

awllower
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  • Your proof uses same indices $u_i u_i$ but OP seeks different indices $ u_i u_j\ \ \ $ – Bill Dubuque Feb 26 '20 at 02:45
  • Sorry for the mistake. It should be fixed now. – awllower Feb 26 '20 at 03:17
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    Thank you for the help. – G.May Feb 26 '20 at 03:45
  • This is really helpful but why is it guaranteed that if $m = \prod_k p_k^{n_k} | u_i^2$ then $\prod_k p_k^{\lceil \frac{n_k}{2}\rceil} | u_i$ – G.May Feb 26 '20 at 09:12
  • @G.May Write the prime factorization of $u_1$ as $u_1=\prod_kp_k^{\ell_k}$. Then $\prod_kp_k^{n_k}\mid u_1^2$ is equivalent with $n_k\leq2\ell_k,,\forall k$. So $\lceil\frac{n_k}2\rceil\leq\ell_k$, and hence $$\prod_kp_k^{\lceil\frac{n_k}2\rceil}\mid u_1.$$ – awllower Feb 26 '20 at 10:08