Diophantine Quartic

Given $y^{2} = x^{4}+2x+13$ . Since x and y are positive integers we have $y^{2} > x^{4}$ $y > x^{2}$ . $y \ge x^{2}+1$ $y^2 \ge x^{4}+2x^{2}+1$ $x^{4}+2x+13 \ge x^{4} +2x^{2}+1$ $2x^{2}-2x-12 \le 0$ $x^{2}-x-6 \le 0$ hence we have $-2 \le x \le 3$ Hence there are 3 possible values of $x$ ,i.e, $1,2,3$ . Of which we see that $x = 1$ and $x = 3$ only gives an integral value of $y$ . So the possible solutions are $(1,4)$ and $(3,10)$ .

So the sum is $1+4+3+10 = 18$ .

First, let us note that $y=x^2$ is a strict lower bound of the solutions to the quartic: $\begin{aligned}x^4+2x+13 &> (x^2)^2 \\ x^4+2x+13 &> x^4 \\ 2x+13 &> 0 \\ &\text{Which is always true.}\end{aligned}$

Also, $y=x^2+4$ is a strict upper bound: $\begin{aligned}x^4+2x+13 &< (x^2+4)^2\\ x^4+2x+13 &< x^4+8x^2+16\\ 2x+13 &< 8x^2+16 \\ 0&< 8x^2-2x+3 \\ &\text{Which is always true.} \\\end{aligned}$

Therefore, $y=x^2+1$ ( Case 1 ), $y=x^2+2$ ( Case 2 ), or $y=x^2+3$ ( Case 3 ).

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Case 1:

$\begin{aligned}x^4+2x+13 &= (x^2+1)^2 \\ x^4+2x+13 &= x^4+2x^2+1 \\ 2x+13 &= 2x^2+1 \\ x^2-x-6 &=0 \\(x-3)(x+2) &= 0\\ x &=3 \\ y&= 10\end{aligned}$

Case 2:

$\begin{aligned}x^4+2x+13 &= (x^2+2)^2 \\ x^4+2x+13 &= x^4+4x^2+4 \\ 2x+13 &= 4x^2+4 \\ 4x^2-2x-9 &=0 \\ &\text{With no integer solutions.}\end{aligned}$

Case 3:

$\begin{aligned}x^4+2x+13 &= (x^2+3)^2 \\ x^4+2x+13 &= x^4+6x^2+9 \\ 2x+13 &= 6x^2+9 \\ 3x^2-x-2 &=0 \\(x-1)(3x+2) &= 0\\ x &=1 \\ y&= 4\end{aligned}$

Therefore, we have that $(x_1,y_1)=(3,10)$ and $(x_2,y_2)=(1,4)$ . Our answer is therefore $3+10+1+4=\boxed{18}$ .

Honestly, it's not too hard to just bash out. It's easy to find the first two solutions, which are just $(1,4)$ and $(3, 10)$ . My favorite part of my (quite brief) solution is proving that none other pairs are possible. Just bear with me, and it's really not too hard to understand. I have a terrible grasp on inequalities like Cauchy-Schwarz and AM-GM and stuff like that. In fact, if anybody wants to help me out there, it'd be most appreciated. Anyways, start off by noticing that $x^4$ is a perfect square. Now recall that the difference between any consecutive squares is simply their sum. For example, the difference between $15^2$ and $14^2$ is just $15+14=29$ . We can use this to our advantage! The reason being is that both $y^2$ and $x^4$ are perfect squares. Because $y^2$ is obviously larger than $x^4$ , we can assume that they are either consecutive squares, or $y^2$ is some other amount larger than $x^4$ . So, bear with me: let's assume that $x^4$ and $y^2$ are indeed consecutive squares. Then their difference is simply $x^2+y$ . Remember that $y$ is greater than $x$ regardless, so we can substitute this in: $x^2+x$ AT LEAST. Remember, it could be $+2x$ or $+3x$ or whatever. But we'll consider the bare minimum as to not skip any cases. Now we have an equation, since we already know that the difference between the squares is $2x+13$ . Let's set it up (remember that we want $2x+13$ to be greater than or equal to $x^2+x$ ): $x^2-x<=13$ . The largest integral value of $x$ that can satisfy this equation is four. So we only need to consider cases less than four! But first let me give an example as to why my solution is valid: let's consider $x=5$ . We will see that plugging this in won't quite reach the next highest square. This is $25^2+2(10)+13 \Longrightarrow 648$ which is less than $26^2$ by twenty-eight. We can prove that their difference diverges (gets larger and larger) as $x$ increases. Let's try $x=6$ . Plugging this in for $x$ produces one thousand three hundred twenty one, which is forty-eight less than $37^2$ . So as we can see, we only need to consider four cases because the larger $x$ gets, the farther away it becomes from the next perfect square. Thus, the desired quantity is $1+3+4+10 \Longrightarrow \boxed{18}$ . :D