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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 $\sum(x_k-1)^4$

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I TeXified part of it, but I'm not sure I got the last part right. Please check. TeXifying the first equation would be easy, too (look at the changes I made), but I'm not sure it is needed, as you seem to have found so many sub/superscripts :-/ – Jyrki Lahtonen Sep 7 '13 at 5:27
@zyx: Don't ask me :-) I think that at least some of them (1,2,3,4?) are included in some font, but I'm not sure. Never bothered to find out. – Jyrki Lahtonen Sep 7 '13 at 6:40
@JyrkiLahtonen, thanks (from another person who had never bothered to find out). If there are fonts that include only a subset of decimal digits as super/subscripts, that is really interesting and unanticipated information. – zyx Sep 9 '13 at 10:52
@zyx: At least I've seen fonts that include symbols $1/2$, $1/3$, $2/3$, $1/4$ and $3/4$. I may be confusing the two... – Jyrki Lahtonen Sep 9 '13 at 11:08

$f(x) = 4x^4 - 7x^3$ 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.

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mind elaborating it a little. or is there any alternative?? – priya Sep 7 '13 at 11:46
The first sentence follows from factorizing the polynomial. The second from $\sum f(x_i) \geq f(x_1) + 13 f(1)$ since $f(1)$ is the minimum and we have $14$ terms. – zyx Sep 7 '13 at 13:18

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 $f(0)=0$).


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.

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