Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\sum(x_k-1)^4$

share|improve this question
1  
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
add comment

2 Answers

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.

share|improve this answer
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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