AAPL wrote:Veritas Prep
If m and n are are positive integers such that m > n, what is the remainder when m^2-n^2 is divided by 21?
1) The remainder when (m+n) is divided by 21 is 1.
2) The remainder when (m-n) is divided by 21 is 1.
Important: the solution presented below assumes a reasonable quantitative maturity.
(If you do not have that, I recommend the excellent step-by-step presentation offered by Brent above.)
$$\eqalign{
& m > n \ge 1\,\,\,{\rm{ints}}\,\,\,\left( * \right) \cr
& {m^{\rm{2}}} - {n^2} = 21K + R \cr
& K,R\,\,{\rm{ints}}\,\,,\,\,\,0 \le R \le 20 \cr
& ? = R \cr} $$
$$\eqalign{
& \left( 1 \right)\,\,m + n = 21J + 1\,\,,\,\,\,J\mathop \ge \limits^{\left( * \right)} 1\,\,\,{\mathop{\rm int}} \cr
& \left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {21,1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = {21^2} - {1^2}\,\,\, \Rightarrow \,\,\,\,R = \,\,20 \hfill \cr
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {20,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = {20^2} - {2^2} = \left( {20 - 2} \right)\left( {20 + 2} \right) = \left( {21 - 3} \right)\left( {21 + 1} \right) = 21\left( {21 + 1 - 3} \right) - 3\,\,\, \Rightarrow \,\,\,\,R = \,\,18\, \hfill \cr} \right. \cr} $$
$$\eqalign{
& \left( 2 \right)\,\,m - n = 21L + 1\,\,,\,\,\,L\mathop \ge \limits^{\left( * \right)} 0\,\,\,{\mathop{\rm int}} \, \cr
& \left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = 3\,\,\, \Rightarrow \,\,\,\,R = \,\,3 \hfill \cr
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {3,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m^{\rm{2}}} - {n^2} = 5\,\,\, \Rightarrow \,\,\,\,R = \,5\, \hfill \cr} \right. \cr} $$
$$\left( {1 + 2} \right)\,\,\,\,\left( {21J + 1} \right)\left( {21L + 1} \right) = 21\left( {21JL + J + L} \right) + 1\,\,\, \Rightarrow \,\,\,\,R = \,\,1\,\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
