# Integers represented by the polynomial

Can every sufficiently large integer be written in the form $a^{100} + b^{101} + c^{102} + d^{103} + e^{104}$ for some non-negative integers $a$, $b$, $c$, $d$ and $e$? I'm only 15 so if u could please write as elemntary as you can! I know that this problem can be solved elementary :)

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No, it is not possible. Given a large integer $N$, there are $N^{\frac 1{100}}$ smaller numbers of the form $a^{100}$. There are even fewer of the forms with higher exponents. This means you can express less than $(N^{\frac 1{100}})^5=N^{\frac 1{20}} \lt N$ numbers this way.

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Hey, i don't really get it why are there there are $N^{\frac 1{100}}$ smaller numbers of the form $a^{100}$ and why there are even fewer of the forms with higher exponents :( –  user51650 Dec 3 '12 at 21:57
@user51650: Let's use squares. Given a large $N$, how many integers of the form $a^2$ are less than $N$?. We must have $a^2 \lt N,$ or $a \lt \sqrt N$. Then if you ask for cubes, it is the cube root of $N$. The argument is the same with higher exponents. –  Ross Millikan Dec 3 '12 at 22:06
@user51650: My argument and Thomas Andrews' are essentially the same. He is more careful than I was about the boundary conditions. In the square example, there are 11 numbers less than 101 of the form a^2 (a can range from 1 through 10), while my square root would give 10. –  Ross Millikan Dec 3 '12 at 22:12
Then if you ask for cubes, it is the cube root of N - i dont get it :( ? –  user51650 Dec 3 '12 at 22:28
@user51650: How many cubes are less than $1000?$ It is 10(OK-11, if you use less than or equal), which is the cube root of 1000. –  Ross Millikan Dec 3 '12 at 22:32

Let $A=\{n\in\mathbb N: \exists a,b,c,d,e\in\mathbb N: n= a^{100}+b^{101}+c^{102}+d^{103}+e^{104}\}$. For $N>0$, let $A_N=\{n<N: n\in A\}$.

If your statement were true, we'd have, amongst other things, that $\lim_{N\to\infty} |A_N|/N =1$.

Now write $n=1+\lfloor\sqrt[100]N\rfloor$. Then the set of $5$-tuples $(a,b,c,d,e)$ such that $a^{100}+b^{101}+c^{102}+d^{103}+e^{104}<N$ is at most $n^5$. Therefore, $$|A_N|<(1+\lfloor \sqrt[100]N\rfloor)^5$$

I think it is therefore pretty obvious that $|A_N|/N\to 0$ as $N\to\infty$.

This is a much stronger result than negating your result - it says that $|A_N|=O(\sqrt[20]N)$, which is a very sparse set.

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