For a a∈Z, if √a∈Q, then √a∈Z. Main idea of this question is this.
Now we'll denote the integer part of a real number x with [x]. So naturally we have x=[x]+{x}So if {x}={x2}, then we have x−[x]=x2−[x2]→x2−x=[x2]−[x]=c∈Z.
So we have x2=x+c, c∈Z. Now same thing can be applied to the pair (x,xn), so we would get xn−x∈Z.
But since we have x2=x+c, we can replace x2 by x+c wherever we want. Looking back to xn−x, we see that we can reduce its degree by applying that so doing this until we have no other choice but to stop, the expression xn−x would have written in a form like bx+d, b,d∈Z, since all coefficents of xn−x always will be integers. So what I am saying is there must be b,d∈Z, such that xn−x=bx+d∈Zbx+d,b,d∈Z⇒x∈Q.
Also going back, we can solve x2−x−c=0, to get exact value of x, and it is 1±√1+4c2This is a rational number, so √1+4c is a rational number but our first observation tells us it must be an integer too. 1+4c is an odd number so √1+4c is odd too. So
2∣1+√4c+1,2x=1+√1+4cThese basicly tell us that x is an integer.