SOLUTION: Proposition: If 'a' is a type 1 unteger and 'b' is a type 2 integer, then (a^2-b) is a type 2 integer.
Note:a type 1 integer is defined as x=3y+1 and i type two integer is defin
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Question 517510: Proposition: If 'a' is a type 1 unteger and 'b' is a type 2 integer, then (a^2-b) is a type 2 integer.
Note:a type 1 integer is defined as x=3y+1 and i type two integer is defined as x=3y+2
This is what have so far:
Proof: We let a be a type 1 integer and b be a type 2 integer. We will show that a^2-b is a type 2 integer. By the definitions of a type 1 integer and type 2 integer, there are integers m and n such that a=3m+1 and b=3n+2. By substitution and the use of algebra we see that a^2-b = (3m+1)2-(3n+2)
= (9m2+6m+1-3n-2)
= (9m2+6m-3n)-1-2
= 3(3m2+2m-n)-1
I know that 3(3m2+2m-n)-1 is a type two integer, I just don't know how to re-write it so it satisfies the definition of a type 2 integer, meaning I don't know how to get it into the x=3y+2 format.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
I presume by your definition that "x is a type 1 integer if x = 3y + 1 for integer y" and similarly for b. There are two ways to show this (the second solution is much easier and more elegant).
Solution 1: Simply expand the expression a^2 - b, e.g.
^2 - (3n+2))
 - 1)
We can write it in the form of a "type two" integer by adding 3 then subtracting 3:
 - 3 + 3 - 1)
The -3 can be factored along with 3(3m^2 + 2m - n):
 + 3 - 1 = 3(m^2 + 2m - n - 1) + 2)
, hence it is a type 2 integer. ∎
--------------------
Solution 2: This involves modular arithmetic (http://en.wikipedia.org/wiki/Modular_arithmetic). It is clear that all integers congruent to 1 mod 3 are type 1 integers, and all integers congruent to 2 mod 3 are type 2 integers. Also,
, b \equiv 2 (mod 3))
so it follows that
)
Therefore a^2 - b is a type 2 integer. ∎
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