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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\

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elcharly64

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 [tex]a^2[/tex] is a type 1 integer

Expanding

[tex]a^2=(3n+1)^2=9n^2+6n+1[/tex]

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

[tex]a^2=3m+1[/tex]

Equating

[tex]a^2=3m+1=9n^2+6n+1[/tex]

solving for m

[tex]m=3n^2+2n[/tex]

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

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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 Β is a type 1 integer

Expanding

If Β 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