equation with Integer part

Let x=y^2

Then we have y^2 = 6 [y]+1. Note that y² is an integer and y≥1

[y] \le (y+1) \Rightarrow y^2 \le 6 (y+1)+1 \Rightarrow y^2-6y-7 \le 0 \Rightarrow y \le 7

Also, [y] \ge (y-1) \Rightarrow y^2 \ge 6 (y-1)+1 \Rightarrow y^2-6y+5 \ge 0 \Rightarrow y \ge 5

So [y] can be 5,6, or 7.

For [y]=5, we get x=31; [y]=6, gives x=37 and [y]=7 gives x=43 which does not satisfy the eqn.

So 31, 37 are the solutions

Thanks Sir (Brilliant Method Sir).

I have solved very Lengthy way.