From an inequality to solving equations

Simplifying $A,B,C$ , we have:

$A=x^4+8x^3+26x^2+40x+24 = (x^4+16x^2+16+8x^3+8x^2+32x) + (2x^2+8x+8)$

$=(x^2+4x+4)^2+2(x^2+4x+4) = (x^2+4x+4)(x^2+4x+6)$

$B=y^4+4y^3+8y^2+8y+3 = (y^4+4y^2+1+4y^3+2y^2+4y) + (2y^2+4y+2)$

$=(y^2+2y+1)^2+2(y^2+2y+1) = (y^2+2y+1)(y^2+2y+3)$

$C=x^4+8x^3+30x^2+56x+48 = (x^4+16x^2+36+8x^3+12x^2+48x) + (2x^2+8x+12)$

$=(x^2+4x+6)^2+2(x^2+4x+6) = (x^2+4x+6)(x^2+4x+8)$

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Assuming there are integers $a,b$ such that $a=x^2+4x+4,b=y^2+2y+1$ , we have:

$A<B<C \Leftrightarrow a(a+2) < b(b+2) < (a+2)(a+4)$

$a(a+2) < b(b+2) \Leftrightarrow a^2+2a < b^2+2b \Rightarrow a^2-b^2+2a-2b < 0 \Leftrightarrow (a-b)(a+b+2)<0$

$b(b+2) < (a+2)(a+4) \Leftrightarrow b^2+2b < (a+2)^2+2(a+2) \Rightarrow b^2 - (a+2)^2 + 2b - 2(a+2) < 0 \Leftrightarrow (b-a-2)(a+b+4)<0$

We have to solve the inequality $\left\{\begin{matrix} (a-b)(a+b+2) < 0\\ (b-a-2)(a+b+4) < 0 \end{matrix}\right.$

Examining each case will help us solve this inequality.

$\text{Case 1: } a+b+2 < -2 (< 0) \Leftrightarrow a+b+4 < 0$

$\Rightarrow \left\{\begin{matrix} a-b > 0\\ b-a-2 > 0 \Leftrightarrow a-b < -2 \end{matrix}\right.$

This is clearly a contradiction since there is no such pair of $(a,b)$ satisfying $0 < a-b < -2$ .

$\text{Case 2: } a+b+2 =-2 \Leftrightarrow a+b+4 =0$ (Contradiction, since $0<0$ is false.)

$\text{Case 3: } a+b+2 =-1 \Leftrightarrow \left\{\begin{matrix} a+3=-b\\ a+1=-(b+2) \end{matrix}\right. \Rightarrow (a+1)(a+3)=b(b+2)$

$\text{Case 4: } a+b+2 >0 \Leftrightarrow a+b+4 >0 \Rightarrow \left\{\begin{matrix} a-b < 0 \Leftrightarrow a <b\\ b-a-2 < 0 \Leftrightarrow b<a+2 \end{matrix}\right. \Leftrightarrow a < b < a+2$

$\Leftrightarrow b=a+1 \Rightarrow b(b+2)=(a+1)(a+3)$

Examining all cases, we can conclude that $(a+1)(a+3) = b(b+2)$ .

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Substitute $a=x^2+4x+4,b=y^2+2y+1$ in $(a+1)(a+3) = b(b+2)$ , we have:

$(x^2+4x+5)(x^2+4x+7) = (y^2+2y+1)(y^2+2y+3)$

$\Leftrightarrow (x^2+4x+5)^2 + 2(x^2+4x+5) = (y^2+2y+1)^2 + 2(y^2+2y+1)$

$\Leftrightarrow [(y^2+2y+1)^2 + 2(y^2+2y+1) + 1] - [(x^2+4x+5)^2 + 2(x^2+4x+5)+1] = 0$

$\Leftrightarrow (y^2+2y+2)^2 - (x^2+4x+6)^2 = 0$

$\Leftrightarrow (y^2+2y+2 - x^2-4x-6)(y^2+2y+2+x^2+4x+6) = 0$

$\Leftrightarrow [(y+1)^2 - (x+2)^2-1][(y+1)^2+(x+2)^2 +3] = 0$

As $(y+1)^2+(x+2)^2 +3 \geq 3 > 0$ , $(y+1)^2 - (x+2)^2-1 = 0$

$\Leftrightarrow (y+x+3)(y-x-1) = 1$

Solving the equation gives us 2 solutions: $(x,y) = (-2,-2), (-2,0)$

Substitute these results in the original problem, we can see that these 2 pairs satisfy the conditions.

Hence there are $\boxed{2}$ different solutions, which are $(x,y) = (-2,-2), (-2,0)$

Let $X = (x+2)^2,$ $Y = (y+1)^2.$ So $X$ and $Y$ are nonnegative, and the inequalities in the problem become $X^2+2X < Y^2 + 2Y < X^2+6X+8.$ That is, $X^2+2X < Y^2 + 2Y < (X+2)^2 + 2(X+2).$ Since $f(x) = x^2+2x$ is an increasing function on the nonnegative real numbers, this implies that $X < Y < X+2.$ So $Y = X+1.$

This means $(y+1)^2 - (x+2)^2 = 1,$ and the only integral squares that differ by 1 are $0$ and $1,$ so $(y+1)^2 = 1$ and $(x+2)^2 = 0,$ which leads to the $\fbox{2}$ solutions $(-2,0),(-2,-2).$

Let us first factorize $A$ , $B$ , and $C$ .

$\begin{aligned} A(x) & = x^4 + 8x^3 + 26x^2 + 40x + 24 \\ & = x^4 + 8x^3 + 24x^2 + 32x + 16 + 2x^2 + 8x + 8 \\ & =(x+2)^4 + 2(x+2)^2 \end{aligned}$

We note that $A(x)$ is symmetrical about $x=-2$ , where it has its minimum value of $0$ .

$\begin{aligned} B(y) & = y^4 + 4y^3 + 8y^2 + 8y + 3 \\ & = y^4 + 4y^3 + 6y^2 + 4y + 1 + 2y^2 + 4y+ 2 \\ & = (y+1)^4 + 2(y+1)^2 \end{aligned}$

We note that $B(y)$ has the same form as $A(x)$ but symmetrical about $y=-1$ , where it has its minimum value of $0$ . In fact $B(x) = A(x-1)$ and all the integer values of $A(x)$ and $B(y)$ are the same.

$\begin{aligned} C(x) & = x^4 + 8x^3 + 30x^2 + 56x + 48 \\ & = x^4 + 8x^3 + 26x^2 + 40x + 24 + 4x^2 + 16x + 24 \\ & = A(x) + 4(x+2)^2 + 8 \end{aligned}$

Again we find $C(x)$ is symmetrical about $x=-2$ , where it has its minimum value of \8). $C(x) > A(x$ ) for all $x$ and the difference $4(x+2)^2 + 8$ increases with $|x+2|$ .

Plotting the graphs of $A(x)$ , $B(x)$ , and $C(x)$ , we find that only $A(-2) = 0 < B(-2) = B(0) = 3 < C(0) = 8$ satisfy the condition. Therefore there are $\boxed 2$ solution pairs $(0,-2)$ and $(0,0)$ .