Question

We say that an integer a is a type 0 integer if there exists an integer n such that a = 3n. An integer a is a type 1 integer if there exists an integer n such that a = 3n + 1. An integer a is a type 2 integer if there exists an integer n such that a = 3n + 2. Prove that if a is a type 1 integer, then a 2 is a type 1 integer\

Answer 1

Answer:

Proof below

Step-by-step explanation:

Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove $a^2$ is a type 1 integer

Expanding

$a^2=(3n+1)^2=9n^2+6n+1$

If $a^2$ is a type 1 integer, then we should be able to find an integer m such as

$a^2=3m+1$

Equating

$a^2=3m+1=9n^2+6n+1$

solving for m

$m=3n^2+2n$

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven

Answer 2

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Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove  is a type 1 integer

Expanding

If  is a type 1 integer, then we should be able to find an integer m such as

Equating

solving for m

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven