Why does the equation $a^4+3b^4=c^4+3d^4$ have solutions in such small integers?

Question

The smallest non-trivial solution of the diophantine equation:

$$a^4+b^4=c^4+d^4\qquad(1)$$

is $(a,b,c,d)=(59,158,133,134)$ (see here, equation 116). One might expect that integers of the form $m^4+3n^4$ are even more sparsely distributed than those of the form $m^4+n^4$. So it seems rather surprising that the equation:

$$a^4+3b^4=c^4+3d^4\qquad(2)$$

has solutions in very small integers, namely, $(a,b,c,d)=(2,3,4,1)$ and $(a,b,c,d)=(7,8,11,2)$.

Question: Is there any reason, eg a simple parametric solution, why equation (2) should have these much smaller solutions, or is this just a random fact?

Addendum

My attempt above in terms of relative sparsity to say why it is surprising that there are solutions of (2) in very small integers is incorrect as pointed out in nickgard’s comment. Taking $1000$ as a threshold, for example, there are 18 pairs $m,n$ with $m^4+3n^4\leq1000$ ($m=1,\cdots,5$ with $n=1,\cdots,3$ and $m=4$ with $n=1,\cdots,3$) but only 14 such essentially distinct pairs with $m^4+n^4\leq1000$ ($m=1,\cdots,4$ with $n=1,\cdots,m$ and $m=5$ with $n=1,\cdots,4$. Similarly, taking $20000$ as a threshold there are 89 pairs for $m^4+3n^4$ but only 63 for $m^4+n^4$.

However, suppose we assume – I know it isn’t exactly correct but it seems a useful first approximation – that the integers $m^4+3n^4$ corresponding to the 18 pairs $m,n$ are randomly distributed over the interval $[1,1000]$. Then the probability that at least one pair of these integers are equal is (the method here is the same as in the birthday problem):

$$1 - \Bigg(\dfrac{1000}{1000}\times\dfrac{1000-1}{1000}\times\dfrac{1000-2}{1000}\times\cdots\times\dfrac{1000-18+1}{1000}\Bigg)\approx 0.14$$

Despite this low probability, there is the one case $1^4+3(3^4)=4^4+3(1^4)=259$.

Making a similar calculation for the 89 pairs in the interval $[1,20000]$, the calculation is:

$$1 - \Bigg(\dfrac{20000}{20000}\times\dfrac{20000-1}{20000}\times\dfrac{20000-2}{20000}\times\cdots\times\dfrac{20000-89+1}{20000}\Bigg)\approx 0.18$$

Note that $0.18$ is the probability of at least one pair of these integers are equal, but despite this low probability there are two equal pairs: $1^4+3(3^4)=4^4+3(1^4)=259$ and $7^4+3(8^4)=11^4+3(2^4)=14689$.

Answer 1

It is true that the integers of the form $m^4+3n^4$ are more sparsely distributed than those of the form $m^4+n^4$. But, the equations are not dealing with distribution, they are dealing with an equality.

Actually, the first equation is equivalent to \begin{equation*} \frac{a^4-c^4}{d^4-b^4}=1 \end{equation*} whereas the second one is equivalent to \begin{equation*} \frac{a^4-c^4}{d^4-b^4}=3 \end{equation*}

Does the second one look any more complicated than the first one now? It's just that the fraction in the LHS becomes $3$ for smaller values and $1$ for larger values.

Is that okay?

Answer 2

Answering my own question following further research.

$$a^4 + b^4 = c^4 + d^4\qquad(1)$$

$$a^4 + 3b^4 = c^4 + 3d^4\qquad(2)$$

Whereas (1) has no non-trivial solutions with $a,b,c,d < 100$ and $gcd(a,b,c,d) = 1$, my search re (2) found three such solutions, namely $(a,b,c,d) = (2,3,4,1),\,(7,8,11,2),\,(23,27,37,1)$.

A property of both equations is that solutions can be used to derive further solutions. It is convenient here to generalise and rearrange (1) and (2) as:

$$a^4 + kb^4 - c^4 - kd^4 = 0\qquad(3)$$

Suppose $(p_1,q_1,r_1,s_1)$ and $(p_2,q_2,r_2,s_2)$ are solutions in integers of (3) for some given $k$. Then $(p_1x+p_2,q_1x+q_2,r_1x+r_2,s_1x+s_2)$ will also satisfy (3) provided that $x$ is chosen such that, on substituting into (3) and expanding the fourth powers, the sum of the terms in $x$, $x^2$ and $x^3$ is zero, that is:

$$4x^3[p_1^3p_2 + kq_1^3q_2 – r_1^3r_2 – ks_1^3s_2] + 6x^2[p_1^2p_2^2 + kq_1^2q_2^2 – r_1^2r_2^2 – ks_1^2s_2^2]+ 4x[p_1p_2^3 + kq_1q_2^2 – r_1r_2^3 – ks_1s_2^3] = 0\quad(4)$$

To see why is this so, note that on expanding the fourth powers, the terms without $x$ will sum to zero because $(p_1,q_1,r_1,s_1)$ is a solution, the terms in $x^4$ will sum to zero because $(p_2,q_2,r_2,s_2)$ is a solution, and (4) ensures that all the other terms sum to zero.

Although (4) is solved by $x = 0$, this is uninteresting since it just leads back to the solution $(p_2,q_2,r_2,s_2)$. To derive a new solution, we want $x\neq 0$, so we can divide (4) by $x$ obtaining a quadratic equation in $x$. If our initial two solutions are selected at random, then we may well find that the values of $x$ implied by that equation are irrational or complex. What we want is for $x$ to be rational, in which case application of a suitable multiplier to $(p_1x+p_2,q_1x+q_2,r_1x+r_2,s_1x+s_2)$ will yield a solution of (3) in integers.

There are several devices that can help select pairs of initial solutions (one of which may be trivial) which will lead to integer or rational values of $x$:

a) Since (3) is an equation in an even power, we can change the sign of any term in a solution and still have a solution.

b) If we can find a solution $(p_1,q_1,r_1,s_1)$ of (3) which is also a solution when the exponent is $1$ rather than $4$, that is, which also solves $a + kb - c – kd = 0$, then we can put $(p_2,q_2,r_2,s_2) = (1,1,1,1)$ and then in (4) the term in x will be zero, so (4) will have a rational solution.

c) Given any non-trivial solution $(p_1,q_1,r_1,s_1)$, we can choose $(p_2,q_2,r_2,s_2) = (u,v,u,v)$ where $u,v$ are integers such that:

$$\dfrac{v}{u} = -\dfrac{p_1^3 – r_1^3}{k(q_1^3 – s_1^3)}\qquad(5)$$

This ensures that the $x^3$ term in (4) will be zero, again leading to a rational solution for $x$.

For the case $k = 3$, the following derivations starting from the solution $(2,3,4,1)$ use all the above devices.

Firstly, changing $2$ to $-2$ we have $(-2) + (3 \times 3) – 4 – (3 \times 1) = 0$, so we can use device (b). Substituting into (4) we have:

$$4x^3[(-2)^3 + (3 \times 3^3) – 4^3 – (3 \times 1^3)] + 6x^2[(-2)^2 + (3 \times 3^2) – 4^2 – (3 \times1^2)] + 4x[0] = 0\qquad(6)$$

$$24x + 72 = 0$$ $$x = -3$$

This yields the solution $((-2 \times -3) + 1,(3 \times -3) + 1, (4 \times -3) + 1, (1 \times -3) + 1)$ which simplifies to $(7,-8,-11,-2)$ from which, changing signs, we have the solution $\bf{(7,8,11,2)}$.

Secondly, changing $1$ to $-1$ and using device (c), we have:

$$\dfrac{v}{u} = -\dfrac{2^3 - 4^3}{3((3^3) - (-1)^3} = \dfrac{56}{84} = \dfrac{2}{3}$$

Putting $p_2 = r_2 = 3$ and $q_2 = s_2 = 2$ and substituting into (4):

$$4x^3[0] + 6x^2[(2^2 \times 3^2) + (3 \times 3^2 \times 2^2) – (4^2 \times 3^2) – (3 \times (-1)^2 \times 2^2)] + 4x^2 [(2 \times 3^3) + (3 \times 3 \times 2^3 – (4 \times 3^3) – (3 \times (-1) \times 2^3)] = 0\qquad(7)$$

$$-72x + 168 = 0$$

$$x = 7/3$$

This yields the solution $((2 \times 7/3) + 3, (3 \times 7/3) + 2, (4 \times 7/3) + 3, (-1 \times 7/3) + 2)$ which simplifies to $(23/3, 9, 37/3, -1/3)$ from which, multiplying by $3$ and changing a sign, we have the solution $\bf{(23,27,37,1)}$.

So the solution $(2,3,4,1)$ implies the other two small solutions of (2). If equation (1) had had a solution in very small integers, similar calculations would probably yield other small solutions.

To show in turn that the existence of the solution $(2,3,4,1)$ is not just a random fact, I offer the following. It is easy to show that if $n$ is an odd integer, then $n^4 \equiv 1 \pmod{2^4}$. Hence if $m$ and $n$ are any odd integers, $m^4 - n^4 = g(2^4)$ for some integer $g$. Substituting the two smallest odd integers $3$ and $1$ we have $g = 5$:

$$3^4 - 1^4 = 5(2^4)\qquad(8)$$

Suppose now that $h$ is an integer such that $5h$ is a difference of fourth powers, say $y^4 – z^4$. In that case we have:

$$h(3^4) – h(1^4) = 5h(2^4) = (y^4 – z^4)2^4 = (2y)^4 – (2z)^4\qquad(9)$$

which can be rearranged as a solution to:

$$a^4 + hb^4 = c^4 + hd^4\qquad(10)$$

But $3$ is such an integer since $5 \times 3 = 2^4 – 1^4$. Substituting $h = 3, y = 2, z = 1$ in (9) we have:

$$3(3^4) – 3(1^4) = 4^4 – 2^4\qquad(11)$$

which can be rearranged as the solution $\bf{(2,3,4,1)}$ to (2).