Is Quartic a square?

$x^4-4x^3+22x^2-36x+18\\ =x^4-4x^3+4x^2+18x^2-36x+18 \\ =(x^2-2x)^2+18(x^2-2x)+18$

Let the expression be equal to $y^2$ , $x^2-2x=a$ where $y\in W$ :

$a^2+18a+18=y^2 \\ a^2+18a+81-63=y^2 \\ (a+9)^2-63=y^2 \\ (a+9-y)(a+9+y)=63$

We can get pairs: $(a+9-y,a+9+y)=(1,63),(3,21),(7,9)$

$\Rightarrow (a,y)=(23,31),(3,9),(-1,1)$

When $a=23 \Rightarrow x^2-2x-23=0$ which does not fetch integer roots.

When $a=3 \Rightarrow x^2-2x=3 \Rightarrow (x-3)(x+1)=0$ which gives us $3$ as one of required roots.

When $a=-1 \Rightarrow x^2-2x=-1 \Rightarrow (x-1)^2=0$ which gives us $1$ as one of the requred roots.

Hence sum $=1+3=\huge\boxed{\color{#3D99F6}{4}}$

Let's find a perfect square $(x^2+bx+c)^2$ that approximates the given polynomial $p(x)$ well. We let $b=-2$ to generate the term $-4x^3$ and then $c=9$ to get the term $22x^2$ . We find that $(x^2-2x+9)^2-p(x)=63$ . It is a straightforward algebra exercise to verify that $(x^2-2x+8)^2$ $<p(x)$ for $x>1+2\sqrt{6}$ . Being trapped between two consecutive perfect squares, $p(x)$ cannot be a perfect square for $x>5.$ We test the cases $x = 1,...,5$ to find that $p(1)=1$ and $p(3)=81$ .

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double i, sum, roota;

for(i=1;i<=1000;i++)

 {

sum = (i*i*i*i) - (4*i*i*i) + (22*i*i) - (36*i) + 18;

roota = Math.sqrt(sum);

if(roota%1==0)

System.out.println(i);

}

Output-

1.0

3.0

Answer - $\boxed{4}$

I remember a really similar problem that just appeared recently: same idea, just bound it between two consecutive perfect squares, then the original expression cannot be a square. Let the given polynomial be $P(x)$ (because I'm lazy), and note that the inequality $P(x)>((x-1)^2-1)^2$ is equivalent to $18(x-1)^2>0$ . The inequality $P(x)<x^4$ is equivalent to $4x^3+36x>22x^2+18$ . Applying AM-GM inequality (or using squares are nonnegative) on the LHS gives $4x^3+36x\ge 24x^2> 22x^2+18$ for all integers $x>3$ . So checking $x=1,2,3$ only gives perfect squares $P(x)=1, 81$ when $x=1, x=3$ , hence the answer is $1+3=\boxed{4}$ .

Let the given expression be denoted by K^2

Then

x^4 - 4 x^3 + 22 x^2 - 36 x + 18 = K^2 ......................................... (1)

We can put the given expression (just to obtain x^4 - 4 x^3) in the form

(x^2 - 2 x + b)^2 + m = K^2 .................................................................(2)

Then

x^4 - 4 x^3 + (4 + 2 b) x^2 - 4 b x + (b^2 + m) = K^2 ..................(3)

Comparing the coefficients of the power of x in (1), (3) , we get

b = 9 , m = 63

Then Eq (2) can be put in the form

(x^2 - 2 x + 9)^2 - K^2 = 63

(x^2 - 2 x + 9 + K)(x^2 - 2 x + 9 - K) = 63

(y + K)(y - K) = 63 , where

y = x^2 - 2 x + 9 ............................................................................................(4)

Then we have

(y + K)(y - K) = 1 X 63 = 63 X 1 = 3 X 21 = 21 X 3 = 7 X 9 = 9 X 7

Then

y + K = 1 , y - K = 63 or y + K = 63 , y - K = 1 .Then y = 32 ......... (5)

OR

y + K = 3 , y - K = 21 or y + K = 21 , y - K = 3 .Then y = 12 ..........(6)

OR

y + K = 7 , y - K = 9 or y + K = 9 , y - K = 7 .Then y = 8 ..................(7)

From (4) , (5) we get

x^2 - 2 x - 23 = 0 ............refused because x is not a positive integer

From (4) , (6) we get

x^2 - 2 x - 3 = 0 , then x = 3 , x = -1 (refused)

From (4) , (7) we get

x^2 - 2 x + 1 = 0 , then x = 1

The sum = 3 + 1 = 4

Solved it like @Nihar Mahajan ..... Just factorized it as (x-1)^4 + 16(x-1)^2 + 1 and then used SFFT..... Great Question btw..!!!