Question 621382
Usually, for problems with consecutive odd numbers, it is enough to call them n and n+2, and the same goes with consecutive even numbers. 
In this case, we have to express the numbers in a way that shows they are odd.
To specify that a number is even, we can call it {{{2m}}} and specify that m is a positive integer, so the even number could be 2, 4, 6, etc.
The numbers right before and right after an even number are consecutive odd integers, so we will use {{{2m-1}}} and {{{2m+1}}} as our consecutive odd integers.
Any pair of consecutive odd integers can be expressed as {{{2m-1}}} and {{{2m+1}}}.
For any pair of consecutive odd integers we can find their m by calculating their average (the even number between them) and dividing by 2.
The product of {{{2m-1}}} and {{{2m+1}}} is
{{{(2m+1)(2m-1)=(2m)^2-1^2=4m^2-1}}}
Since {{{m}}} was a positive integer, {{{m^2}}} is a positive integer and
{{{4m^2=4*m^2}}} is a multiple of 4.
{{{4m^2-1}}} is one less than {{{4m^2}}}, so it is 1 less than a multiple of 4.