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# Show that no square number is of the form of $3n-1.$

Last updated date: 20th Jun 2024
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Hint: In this question, we will use the contradiction method to prove the given statement. First, we will assume that the square of a number is equal to 3n-1. After this, we will use Fermat’s little theorem to contradict the assumption. We will also see what Fermat’s little theorem is?

Complete step-by-step solution:
We will first assume a number N.
And let ${N^2}= 3n-1.$
$\Rightarrow$ ${N^2} +1 = 3n.$
It is clear from the above equation that ${N^2}+1$ is the multiple of 3.
Now, we will rearrange the term ${N^2}+1$ by adding and subtracting 1 as follow:
${N^2}$+1= ${N^2}+1+1-1 = ({N^2}-1)+2.$
Now, let us see the statement for Fermat’s little theorem.
According to this theorem, if ‘a’ is not divisible by ‘p’ where ‘a’ and ‘p’ are integers, then the number ${a^{p - 1}} - 1$ is an integer multiple of ‘p’.
For example – if a= 2 and p = 7, then ${2^6} =64$ and $64 – 1 =63 = 7 \times 9$ is thus a multiple of 7.
So, using this theorem, we can say that $({N^2}-1)$ = ${N^{3 - 1}} - 1$ is divisible by 3.
But in$({N^2}-1)+2$, where we are adding 2, the number is now not divisible by 3.
That is ${N^2}+1$ is not divisible by 3. This contradicts our assumption.
Hence, It is proved that no square number is of the form $3n – 1.$

Note: This type of question is generally solved by contradiction method. In this method, the statement is proved by contradicting the assumption. In this question, you must remember Fermat's little theorem. This theorem can also be stated in other as follow:
If ‘p’ is a prime number then for any integer ‘a’, the number ${a^p}- a$ is an integer multiple of ‘p’.
For example- if $a =2$ and $p =7$ then ${2^7}= 128$, and $128 -2 = 126$, which is an integer multiple of 7.