4
$\begingroup$

Consider the following two constraints on a set of integers: $$a^2+b^2+c^2=d^2+e^2+f^2$$ $$a^4+b^4+c^4=d^4+e^4+f^4$$

There are simple solutions like $a^2=b^2=c^2=d^2=e^2=f^2$, or pairs are the same value. But is it possible for the squares of these variables to all be distinct?

The best I've been able to come up with so far, is that if there are integers $x,y,z$ such that $$\begin{aligned} a^2,\,d^2 &=x+y,\;x-y\\ b^2,\,e^2 &=x-y-z,\;x+y+z\\ c^2,\,f^2 &=x+z,\;x-z \end{aligned}$$

Then there would be a solution. This is encouraging because it reduces the number of variables, however I'm not sure if that's just a dead end, because I can't show that all solutions must have this form, and I don't know how to choose $x,y,z$ such that so many linear combinations of them form squares.

What is a good strategy to approach this problem?

$\endgroup$
  • $\begingroup$ A symmetric polynomial idea. Let $f_{a,b,c}(x)=(x-a^2)(x-b^2)(x-c^2)$. Then $$f(x)=x^3-(a^2+b^2+c^2)x^2+(a^2b^2+a^2c^2+b^2c^2)x - a^2b^2c^2.$$ Now, $a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)$, so your condition requires $f_{a,b,c}(x)=f_{d,e,f}(x)+C$ for some constant integer $C$. $\endgroup$ – Thomas Andrews Oct 22 '15 at 5:55
  • 1
    $\begingroup$ math.stackexchange.com/questions/1037013/… $\endgroup$ – individ Oct 22 '15 at 6:49
1
$\begingroup$

There are numerical solutions, such as $(a,b,c,d,e,f)=(7,7,14,2,11,13)$ or $(20,21,37,12,29,35)$ or $(0,7,7,3,5,8)$. I don't know whether a complete solution has been worked out. Gloden gives a 2-parameter family, $$(2p^3q+5p^2q^2+2pq^3,p^4+2p^3q+7p^2q^2+6pq^3+2q^4,2p^4,2p^3q-2pq^3-2q^4,p^4+2p^3q-p^2q^2-2pq^3,2p^4+2p^3q+p^2q^2+2pq^3+2q^4,4p^3q+6p^2q^2+6pq^3+2q^4)$$

Ramanujan wrote, if $a/b=c/d$ then $$(a+b+c)^r+(b+c+d)^r+(a-d)^r=(c+d+a)^r+(d+a+b)^r+(b-c)^r$$ for both $r=2$ and $r=4$. See the article by M D Hirschhorn, Two of three identities of Ramanujan, The American Mathematical Monthly Vol. 105, No. 1 (Jan., 1998), pp. 52-55.

There is also a discussion of this system of equations at https://sites.google.com/site/tpiezas/014 and on the pages linked from that one.

$\endgroup$
1
$\begingroup$

One strategy to solve,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag1$$

is to use the substitution,

$$(p+qu)^k+(r+su)^k+(t+u)^k = (p-qu)^k+(r-su)^k+(t-u)^k\tag2$$

Expanding and factoring for $k=2,4$, one gets,

$$pq+rs+t = 0\tag3$$

$$p^3q+r^3s+t^3+(pq^3+rs^3+t)u^2 = 0\tag4$$

and is just a quadratic in $u$. Eliminating $t$ between $(3),(4)$, one ends up with a quartic polynomial to be made a square. This has obvious rational points, and is birationally equivalent to an elliptic curve. Hence, the general solution to $(1)$ for $k=2,4$ needs an elliptic curve. A special case is,

I. $x_1+x_2+x_3 = y_1+y_2+y_3 = 0$

This can be seen with the smallest solutions of $(1)$,

$$0^k+7^k+(-7)^k = 3^k+5^k+(-8)^k$$

$$1^k+9^k+(-10)^k = 5^k+6^k+(-11)^k$$

In general,

$$p^k + q^k + (-p-q)^k = r^k + s^k + (-r-s)^k$$

for $k=2,4$, where,

$$p^2+pq+q^2 = r^2+rs+s^2$$

and is easily solved parametrically. This condition is essential to Ramanujan's 6-10-8 identity discussed in this post.

P.S Another special case has $x_1+x_2 = y_1+y_2$ (like Myerson's $20,21,37;12,29,35$) and can be solved by quadratic forms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.