Question

Show that the following equations have no integer solutions: (4) 25 + 12x2 + 24x + 1 = 0. (1) x4 + x + 1 = 0; (2) 12x + 5 = y2; (3) x? +6x +1 = 0;

Answer 1

(1) Suppose is an integer solution to:

2° ++1=0

Then
1= -1(2+1)
, So divides 1.

>> r= 1/-

but 1 or -1 are not solutions to as

2° ++1=0

11+1+1=3
and $(-1)^4-1+1=1$

Hence the equation has no integer solutions

(2)For any integer y we have exactly one of the following hold:

$y\equiv0(\textup{mod}3)$ or $y\equiv1(\textup{mod}3)$ or $y\equiv2(\textup{mod}3)$

$\Rightarrow y^2\equiv0(\textup{mod}3)$ or $y^2\equiv1(\textup{mod}3)$ or $y^2\equiv4\equiv1(\textup{mod}3)$ respectively.

Therefore $y^2$ is 0 or 1 modulo 3.

But for any integer x, $12x+5\equiv 5\equiv 2(\textup{mod}3)$

But $2\not\equiv 0/1(\textup{mod}3)$

$\Rightarrow 12x+5=y^2$ has no integer solution

(3) If x is an integer solution to the equation

$x^7+6x+1=0$

Then $1=-x(x^6+6)$ , So divides 1

>> r= 1/-

But 1 or -1 are not solutions to

$x^7+6x+1=0$ as

$1^7+6\times1+1=8$ and $(-1)^7+6\times-1+1=-6$

Hence the equation has no integer solution

(4) Suppose x is an integer solution to

$x^5+12x^2+24x+1=0$

Then $1=-x(x^4+12x+24)$ , So x divides 1

>> r= 1/-

But 1 or -1 are not solutions to

$x^5+12x^2+24x+1=0$ as

$1^5+12\times1^2+24\times1+1=38$

and $(-1)^5+12\times(-1)^2+24\times(-1)+1=-12$ Hence the equation has no integer solution