A little number theory

A quick way to the answer: if $(m,n)$ is a solution pair, then so is $(-n,-m)$ ; this can easily be checked by substitution. Either this is a new solution pair, in which case the sum of the two solution pairs is zero, or it is the same solution pair; this only happens when $m=-n$ , or put another way, when $m+n=0$ , so again the contribution to the sum of all solution pairs is zero. The final thing to note is that there is at least one solution, $(0,0)$ , so the required sum exists. (Note that this shortcut method only helps to find the answer to the question as stated - the other posted solution and discussion is more complete.)

You don't even need to find any solutions !!!!!!!

It is given that

$m^5 - n^5$ = 16mn

If we assume that certain (x ,y) is a solution to above equation then it satisfies it

Therefore

(x, y) ----> $x^5 - y^5$ = 16xy

Also putting (-y, -x) in the equation gives ,

(-y, -x) ---> $x^5 - y^5$ = 16xy which is same as we got on substituting (x, y)

So whenever (x, y) is a solution then (-y,- x) is also a solution

thus the sum of all (m,n) satisfying the equation will be $\boxed {0}$

If $m=n$ , then $m=n=0$ . Thus $(0,0)$ is a solution.

If $m \neq n$ and $m,n \ge 0$ , we must have $m > n \ge 0$ . Then $16mn \; = \; (m-n)(m^4 + m^3n + m^2n^2 + mn^3 + n^4) \; \ge \; m^4 + m^3n \; > \; m^3n$ so that $m^2 < 16$ . The only possible value combinations of $m,n$ are now $(2,1)$ , $(3,1)$ and $(3,2)$ , and none of these are solutions.

If $m \neq n$ and $m,n \le 0$ , then $(-n)^5 - (-m)^5 = 16(-n)(-m)$ with $-n \neq -m$ and $-n , -m \ge 0$ . As we have just seen, there are no solutions in this case either.

If $m > 0 > n$ then $m^5 + |n|^5 + 16m|n| = 0$ , which is impossible.

If $n > 0 > m$ then, putting $j = -m$ we have $j^5 + n^5 = 16jn$ with $j,n > 0$ . If we put $a = \mathrm{max}(j,n)$ then $a^5 < j^5 + n^5 = 16jn \le 16a^2$ , and hence $a^3 < 16$ , so that $a = 1$ or $2$ . This leaves us with the possibilities $(-1,1)$ , $(-1,2)$ , $(-2,1)$ and $(-2,2)$ for $(m,n)$ , and only the last of these works.

Thus the only solutions are $(0,0)$ and $(-2,2)$ , making the answer $0+0+(-2)+2= \boxed{0}$ .

The plot of x^5-y^5=16xy, and Diophantine solution to this equation using WolframAlpha confirms (2,-2) or (-2,2) as the only answer in addition to (0,0).

Answer=(2-2)=0

$(0,0)$ is the obvious solution.

$GCD(m,n)=1$ , because if $GCD(m,n)=d , d \neq 1$ then $n=dn'$ , $m=dm'$ and

$d^5 \big((m')^5-(n')^5\big)=16 d^2(m')(n') \implies d^3 \big((m')^5+(n')^5\big)=16 m'n'$

$d^3$ cannot be made by any of the terms in $16m'n'$ .

Knowing $GCD(m,n)=1$ , we rewrite

$m^5-16mn=m(m^4-16n)=n^5 \implies m|n^5$

which is not possible, unless $m=1$ . In similar ways, we can deduce $n=1$ . But, they don't lead us to any solution, so we are left with $(0,0)$ only.

First of all we consider the case m=n=0, which is a solution.

In the equation we have m^5 and 16mn have a common divisour m. As m and n are integers we can conclude that n^5 has also a divisour m. In the same way we find that m^5 has a divisour n. This is not possible for any pair of integers (m ; n) so for the case m≠n we have no solutions.

Therefore the only solution is m=n=0 and the answer is 0.

Wrote an AWK script to run check pairs between 1 and 100,000. Nothing was found. That seemed to indicate something ...

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BEGIN {
  totalmn = 0;
  count=0
  trips=0;
  for ( m=2; m<=100000; m++ ) {
    for ( n=1; n<m; n++ ) {
      if ( m <= n ) {
        print "m<=n!: m " m " n " n;
        break;
      }
      if ( ( m^5 - n^5 ) == ( 16*m*n ) ) {
        print "m " m " and n " n " work";
        totalmn+=(m+n);
        count++;
      }
      trips++;
    }
    if ( m % 1000 == 0 ) print m;
  }
  print "found " count " pairs";
  print "total of m's and n's are " totalmn;
  print "total trips through the loops: " trips;
  exit(0);
  }

A simpler way is just plot the graph of the equation $x^{5} - y^{5} = 16xy$ and search over integral values of $x$ and $y$

I plotted the graph using Desmos-

We get $x = -2 ,0$ and $y=2,0$ .

Therefore the sum of all possible values $= 0 - 2 + 0 +2 = 0$

Hence, $ANSWER:\boxed {0}$

Clearly, m and n are both even or both odd. Letting m = n + 2k, and expanding, the left side overwhelms the right side.

From this initial equation, we can get this equation : ( ( $\frac{m^{2}}{n}$ ) - ( $\frac{n^{2}}{m}$ )) x (( $\frac{m^{2}}{n}$ )+( $\frac{n^{2}}{m}$ )) = 16 . We can then assume that ( $\frac{m^{2}}{n}$ ) - ( $\frac{n^{2}}{m}$ ) = a and that (( $\frac{m^{2}}{n}$ )+( $\frac{n^{2}}{m}$ ) = $\frac{16}{a}$ where a is any number. We can take these two equations and make them into one. This gives us : $\frac{2n^{2}}{m}$ = $\frac{a^{2} - 16}{a}$ . If we were to say that m = y and that n = x, we could isolate y. This would give us : y = f(x) = $\frac{2ax^{2}}{a^{2}-16}$ . This is the function of a parabola. We can then see that if if we were to superpose each resultant function with a different value of "a" (where "a" would go from 0 to infinity), then all the XY plane would be part of this superposition. Meaning that the sum of each pair of variable (x,y) or (n,m) has to go to zero.