Question 263542
A way to guarantee that a number will be odd is to call it
{{{2n + 1}}}
Odd integers skip every other number, so I can
call them {{{2n+1}}},{{{2n+3}}},{{{2n+5}}},{{{2n+7}}}
given:
{{{(2n + 1)(2n + 3) = (2n + 5)(2n + 7) - 64}}}
{{{4n^2 + 8n + 3 = 4n^2 + 24n + 35 - 64}}}
{{{8n + 3 = 24n - 29}}}
{{{16n = 32}}}
{{{n = 2}}}
and
{{{2n + 1 = 5}}}
{{{2n + 3 = 7}}}
{{{2n + 5 = 9}}}
{{{2n + 7 = 11}}}
The numbers are 5,7,9, and 11
check:
{{{5*7 = 9*11 -64}}}
{{{35 = 99 - 64}}}
{{{35 = 35}}}
OK