Is it possible to find unequal non-zero integers with their harmonic, geometric, arithmetic, and quadratic means all also being integers?

Question

I searched the internet, and it is easy to find integers $a,b$ that have HM, GM and AM all integers. However, QM seems more difficult.

Answer 1

Including the arithmetic and harmonic means in the problem is somewhat vacuous, since these means are always rational for any integer inputs (apart from the harmonic mean when the inputs are negatives of each other) and may be converted to integers through multiplication by the least common denominator. The real problem is whether the geometric and quadratic means can also be both rational or equivalently both integers. For unequal integer inputs that answer is No.

Assume $x$ and $y$ are unequal, relatively prime integers having an integer QM (a rational QM in this case is not possible except for integers, due to the radicand in the QM having a squarefree denominator). Then they must be of the form

$x=\pm(k^2-2kl-l^2)$

$y=\pm(k^2+2kl-l^2)$

where the squares of these numbers are the first and third terms of the arithmetic progression, the second term in this progression being the square of the QM. For the geometric progression also to be an integer, the product $xy$ must also be a square. Since the expressions inside the $\pm$ signs above differ only in the sign of one term, this product can be rendered as a difference of squares:

$(k^2-l^2)^2-(2kl)^2=\pm m^2$ (**)

If the $\pm$ sign in the above product (**) is $+$, then $m^2$ and $(k^2-l^2)$ are two squares and

$(k^2-l^2)^2+(2kl)^2=(k^2+l^2)^2$

becomes a third square in arithmetic progression, whose common difference is the square $(2mn)^2$. If the sign in (**) is $-$, then $m^2, (2kl)^2, (k^2+l^2)^2$ are similarly in arithmetic progression with the common difference $(k^2-l^2)^2$.

This conclusion is contradicted, because squares in arithmetic progression in fact cannot have a common difference which is also a square. Keith Conrad, in a discussion of these progressions citing References [1,2], finds that the common difference can be $n$ times a perfect square only if the elliptic curve $y^2=x(x+n)(x-n)$ has nonzero rational points, and such solutions exist only for certain whole numbers $n$ the smallest of which is $5$ (the squares $31^2, 41^2, 49^2$ successively differ by $5×12^2$).

The nonexistence of three squares in arithmetic prigression differing by a fouth square may also be proven another way. Let $a^2, b^2, c^2$ be any three such squares in order. If the common difference is $d^2$ then $(b^2-d^2)(b^2+d^2)=b^4-d^4=(ac)^2$. This lacks any solutions with all terms nonzero according to Mordell[3].

So the required existence of squares in arithmetic progression with a square common difference fails, and the simultaneous integer values for GM and QM with unequal integer inputs goes down with it.

We can, however, easily render unequal integer pairs with integer AM, HM and QM (leaving out the GM). Examples include:

$\{x,y\}=\{4,28\}; \text{ AM = 16, HM = 7, QM = 20; minimal positive solution}$

$\{x,y\}=\{-3,21\}; \text{ AM = 9, HM = }-7\text{, QM = 15; minimal absolute value solution}$

Cited References

1. W. A. Coppel, Number Theory: An Introduction to Mathematics. Part B, Springer-Verlag, New York, 2006.

2. N. Koblitz, Introduction to Elliptic Curves and Modular Forms, 2nd ed., Springer–Verlag, New York, 1993.

3. R. J. Mordell, Diophantine Equations (London: Academic Press, 1969), 17-18.