Question

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

Answer 1

Let there exist $N$ such that $N^2=3n-1$

$\Rightarrow N^2+1=3n$

which is clearly a multiple of $3$

$N^2+1=N^2+1+1-1\Rightarrow N^2+1=(N^2-1)+2$

Now using fermats theorem if $N$ is prime to $p$ then $N^{p-1}-1$ is divisible by $p$

$\Rightarrow N^2-1=N^{3-1}-1$ is divisible by $3$

$\Rightarrow N^2+1=(N^2-1)+2$ is not divisblle by $3$ and $2$ can not be divided by $3$

So this contradict our asssumption

Hence proved that no square number is of form $3n-1$