Given an $n$ degree equation in 2 variables ($n$ is a natural number) $$a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n=y^n$$

If all values of $a$ are given rational numbers, are there any known minimum or sufficient conditions for $x$ and $y$ to have:

  1. Real number
  2. Rational number
  3. Integer

solutions and how many of them would exist. If it is not known/possible (or too hard) for an $n$ degree polynomial, do such conditions exist for quadratic ($n=2$) and cubic ($n=3$) polynomials.

  • $\begingroup$ For every positive value of polynomial there exist a real solution to $y$ $\endgroup$
    – kingW3
    Dec 14, 2014 at 13:25
  • $\begingroup$ Why has someone given a close vote on this? I'm just looking for general rules, such as when is the solution an integer and so forth, not a specific answer. $\endgroup$ Dec 14, 2014 at 17:40
  • $\begingroup$ @user45195: Perhaps it is better if the question is edited so the $a_i$ and $x,y$ are just in the rationals. $\endgroup$ Dec 18, 2014 at 1:00
  • $\begingroup$ Yes, I've edited it. $\endgroup$ Dec 21, 2014 at 18:14

3 Answers 3


This addresses user45195's question and is too long for a comment.

When I said too broad, it was because the question originally didn't limit the field of $x$. A familiar field is the complex numbers $\mathbb{C}=a+bi$, of which a special case are the reals $\mathbb{R}$, and even more limited, the rationals $\mathbb{Q}$.

If $x$ was in $\mathbb{C}$, then it's just an old result (Fundamental Theorem of Algebra), that for any $y$, one can always find $n$ roots $x$ that solve the equation in your post, and there's nothing new to be said.

However, if $x,y$ are limited to the rationals $\mathbb{Q}$, that's where it gets interesting. The equation,

$$f(x) = y^2\tag1$$

where the degree $d$ of $f(x)$ is either $d = 2,3,4,5$ has been extensively studied. See algebraic curve, including Pell equations ($d=2$), elliptic curve ($d=3,4$), and hyperelliptic curve ($d=5$). For,

$$f(x) = y^3\tag2$$

then $d=3$ still has special cases as elliptic curves. For $d=4$, see trigonal curve. However, for the higher,

$$f(x) = y^m\tag3$$

where $d,m>3$, is more complicated. See superelliptic curve.

  • $\begingroup$ So, for say a 10 degree polynomial, brute force is the only way to find solutions? $\endgroup$ Dec 21, 2014 at 14:17
  • $\begingroup$ @user45195: Well, the most obvious way. But for lower-deg polys there are alternative methods and also, given an initial rational point, generally an infinite more can be found systematically (not brute). $\endgroup$ Dec 21, 2014 at 16:26
  • $\begingroup$ Thank you, I was asking from a computational point of view. $\endgroup$ Dec 21, 2014 at 18:14

(The OP suggested a connection to this post.)

Because this question is too broad, it spreads thin and may be vague. I suggest it be limited so coefficients $a_i$ are rational, and $x,y$ are also rational.

Having said that, two nice results are discussed in Kevin Brown's website.

I. Deg 2: The sum of $24$ consecutive squares.

$$F(x) = x^2+(x+1)^2+(x+2)^2+\dots+(x+23)^2=y^2\tag1$$

$$F(x) = 24x^2+552x+4324=y^2$$

which has solution,

$$x=p^2+70pq+144q^2,\quad\quad y =10(7p^2+30pq+42q^2)$$

where $p,q$ solve the Pell equation $p^2-6q^2=1$. This has an infinite number of integer solutions with the case $p,q = 1,0$ yielding the famous cannonball stacking problem,

$$1^2+2^2+3^2+\dots+24^2 =70^2$$

II. Deg 3: The sum of $n$ consecutive cubes.

$$G(x) = x^3+(x+1)^3+(x+2)^3+\dots+(x+n-1)^3=y^3\tag2$$

$$G(x) = n x^3 + \tfrac{1}{4}n(n - 1)\big(6x^2 + 4 n x - 2x + n(n - 1)\big) =y^3$$

a solution of which (by Dave Rusin) is,

$$x=\tfrac{1}{6}(v^4 - 3v^3 - 2v^2 + 4),\quad\quad n=v^3$$

hence for $2^3=8$ and $4^3=64$ cubes,

$$(-2)^3+(-1)^3+\dots+3^3+4^3+5^3 = 6^3$$

$$6^3+7^3+8^3+9^3+10^3+\dots+69^3 = 180^3$$

and so on.

III. Deg 4: The sum of 4th powers in arithmetic progression.

No analogous results known so far. See linked post in first line.

  • $\begingroup$ Thank you, but I never mentioned that terms are in an AP. $\endgroup$ Dec 18, 2014 at 9:26
  • $\begingroup$ @user45195: Yes, I had to make assumptions on your $x,y$ because the question is too broad. One can start with special cases, and may then get interesting results. By the way, what field is your $x$ supposed to be in? (I think that omission is the reason for someone's down-vote.) $\endgroup$ Dec 18, 2014 at 17:03
  • $\begingroup$ a) If it is too broad, does that mean there exists no general solution to solve such equations (like the way roots can directly be found for quadratic equations, etc.)? b) I'll edit the question to include only rational coeffecients. c) What do you mean by field? $\endgroup$ Dec 19, 2014 at 11:08

You can determine whether a real solution to that equation exists; obviously, there is always one if $n$ is odd, since the left side, $y^n$ can be chosen, independent of the right side, to equal whatever we want. For even $n$, we want to know whether the left hand polynomial is ever positive - this clearly holds if $a_0$ is positive, since the function would tend to $\infty$ as $|x|$ gets large. However, if $a_0$ is negative, then the left hand function will be non-negative somewhere if and only if it has at least one real root. Thus, the problem would then collapse to determining if a univariate polynomial has a root and, in general, this is possible to accomplish computationally via Sturm sequences.


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