sum of 14 4th powers and sum of 14 cubes Prove that
$4(x_1^4 + x_2^4 + x_3^4 + \dots + x_{14}^4) = 7(x_1^3+ x_2^3 + x_3^3 + \dots + x_{14}^3)$ has no solution in positive integers.
Hint : suppose on the contrary  $\sum_{k=1}^{14} {(x_k^4 - \frac74 x_k^3)} = 0$ . also use $\sum(x_k-1)^4$
 A: $f(x) = 4x^4 - 7x^3$ is positive except for $f(0)=0$ and $f(1)=-3$. Only the smallest positive values, $f(2)=8$ and $f(-1)=11$, are consistent with $\sum f(x_i) \leq 0$ , all other integer $f(x)$ would overwhelm the extreme case  where $13$  of $14$ values are $-3$.  
The problem is a small finite search from this point, and requesting positive solutions leaves only $f(1)$ and $f(2)$ in the game.  
A: Let me elaborate further on zyx's solution.
Since $f(x) = 4x^4 - 7x^3$ is positive at all integer inputs $x \neq 0,1$, the only way to have $\sum_{k=1}^{14}f(x_i) = 0$ is either:
1) $x_i = 0$ for all $i$ (since $f(0)=0$).
or
2) Some number of the $x_i$'s are $1$ (because $f(1)=-3$ is negative and so will cancel out any positive contribution from the other $f$ values).
Possibility $1$ is not valid since we seek positive solutions. That leaves us with possibility $2$.
Suppose $j$ of the $x_i$ values are $1$. Then we are trying to solve $S = 3j$ where $S$ is the sum of the $f(x_m)$ for which $x_m\neq 1$.
Now since $j\leq 14$ it is clear that $0\leq S\leq 42$ and $S$ is a sum of non-negative integers.
Aha! $f$ only takes non-negative values less than $42$ for $x=-1,0,2$ (check this). The values of $f$ at these points are $11,0,8$ respectively.
So we let $a$ be the number of $x_i = -1$ occurrences in $S$, $b$ be the number of $x_i = 0$ occurrences and $c$ be the number of $x_i = 2$ occurrences.
Then we are solving the system:
$S = 11a + 8c = 3j$
$a+b+c = 14-j$
in integers $a,b,c,j$ with $1\leq j \leq 14$ and $0\leq a,b,c\leq 14$. In fact the first equation tells you that $a=0,1,2,3$, $c=0,1,2,3,4,5$ and that $c \equiv -a \bmod 3$. This narrows down the possibilities. You need only check now that in each case the equations cannot be satisfied.
