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Find all $n\in\mathbb Z$ which satisfies $1^n+9^n+10^n=5^n+6^n+11^n$

for $n=2\ or\ n=4$ it is equal but are there other numbers?

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  • $\begingroup$ Modulo $5$ or $3$ says $n$ must be even $\endgroup$ – lab bhattacharjee Nov 21 '13 at 10:49
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    $\begingroup$ $n=0. {}{}{}{}{}$ $\endgroup$ – Casteels Nov 21 '13 at 10:49
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    $\begingroup$ for $n\ge 9$ we have $11^n=(10+1)^n>10^n+n 10^{n-1}>10^n+9^n$, so we will have the right hand side is greater than the left hand side. So you are left with $n<9$. On the other hand, $n$ cannot be negative, because of 11 being a prime number. Therefore, it remains to check $0\le n\le 8$. $\endgroup$ – Tigran Hakobyan Nov 21 '13 at 10:52
  • $\begingroup$ @TigranHakobyan first argument is clear now. but why are negatives not allowed if 11 is prime $\endgroup$ – derivative Nov 21 '13 at 10:58
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    $\begingroup$ Negative is no problem, left side is $\gt 1$, right side is less than $1$. $\endgroup$ – André Nicolas Nov 21 '13 at 11:04
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In general, the Lander-Parkin-Selfridge conjecture states that,

$$x_1^k+x_2^k = y_1^k+y_2^k\tag{1}$$

has no non-trivial rational solutions for $k>4$,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag{2}$$

has none for $k>6$, and so on. Thus, for your question, you need not check $k>6$.

Then again, for bigger $x_i,y_i$, one can always be surprised. For example, we have,

$$15^6 + 18^6 = 19^6 + (-118)^3$$

$$1118^8 + 1937^8 + 2502045^4 = 455^8 + 1846^8 + 3200691^4$$

and an infinite more co-prime solutions like it which almost, but not quite, violate $(1)$ and $(2)$.

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  • $\begingroup$ Admitedly, the conjecture is not yet proven, but Ekl in a 1997 paper has checked $x_i,y_i$ in a relatively high range. $\endgroup$ – Tito Piezas III Nov 30 '14 at 0:24
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You have tried all of $0$ through $4$. If you note that $11^5 \gt 10^5+9^5$ and the discrepancy gets worse with $n$, you know there are no more. For $n$ negative the right side will have a power of $11$ in the denominator and the left will not, so there are no negative solutions. You have them all.

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