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For all $n \in \mathbb{N}$, solve the Diophantine equation $x^n-y^n=1001$, where $x,y \in \mathbb{N}$.

The cases $n=1,2$ are trivial ones. But for $n>2$ I can't find any solutions. How could I prove that there are no integer solutions for $n>2$?

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  • $\begingroup$ Is $n\in\mathbb N$? Are $x,y\in\mathbb N$? $\endgroup$ May 22, 2015 at 16:17
  • $\begingroup$ Yes, $n, x,y \in \mathbb{N}$, I corrected $\endgroup$
    – Leox
    May 22, 2015 at 16:19
  • $\begingroup$ Use the factorization of $x^n - y^n$, perhaps. $\endgroup$
    – Sawarnik
    May 22, 2015 at 16:19
  • $\begingroup$ only factorisation is not enought $\endgroup$
    – Leox
    May 22, 2015 at 16:20
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    $\begingroup$ Totally inelegant, but I observe that brute force seems to suffice for this; cases are limited to $n < 10$ (since $2^{10} = 1024 > 1001+1$. $\endgroup$
    – Brian Tung
    May 22, 2015 at 16:23

2 Answers 2

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Since $x,y \in \mathbb{N}$ and $x^4$ - $y^4$= $(y + h)^4$ - $y^4 = 4y^3h + 6y^2h^2 + 4yh^3 + h^4$ we have x - y = 1 for all n≥4 because $7^4$ = 2041 > 1001 then 7 and greater than 7 candidates (11, 13, 77,…) must be discarded. Therefore ${n\choose1}y^{n-1} +{n\choose2}y^{n-2} + … + {n\choose1}y + 1$ = 1001; hence

y[${n\choose1}y^{n-2} +{n\choose2}y^{n-3} + … + {n\choose1}$] = 1000 = $2^35^3$ and y = 1, 2, 4, 8, 5, 10, 25,….We discard these candidates to factors by the following calculation which shows that 1000 is strictly between the given powers:

$2^9 – 1 < 1000 < 2^{10} – 1$

$3^6 – 2^6 < 1000 < 3^7 – 2^7$

$5^4 – 4^4 < 1000 < 5^5 – 4^5$

$6^4 – 5^4 < 1000 < 6^5 – 5^5$

$9^3 – 8^3 < 1000 < 9^4 – 8^4$

$11^3 – 10^3 < 1000 < 11^4 – 10^4$

$26^2 – 25^2 < 1000 < 26^3 – 25^3$.

This finish the proof for all n >3. The case n = 3 can be solve by a quadratic equation and careful calculation about impossibility of discriminant as a (necessary) square.

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We'll$\let\leq\leqslant\let\geq\geqslant$ use the following, which can easily be proved using for example Bézout's theorem:

If $p$ is prime, then $p\mid x^n-y^n$ implies $p\mid x^{\gcd(n,p-1)}-y^{\gcd(n,p-1)}$.

Note $1001=7\cdot11\cdot13$ and $x^n-y^n=(y+(x-y))^n-y^n\geq(x-y)^n$. If $x-y$ and $n$ are large enough, we might use the inequality $(x-y)^n>1001$.

  • If $n=3$, then $11\mid x^3-y^3$ implies $11\mid x-y$, but $11^3>1001$.
  • If $n=4$, then $77\mid x^4-y^4$ implies $77\mid x^2-y^2$. If $x-y>1$ we would have $x-y\geq7$, but $7^4>1001$. Hence $x-y=1$ and $77\mid x+y$. But then $x^4-y^4=(x-y)\cdot(x^3+\cdots)>x^3\geq34^3>1001$.
  • If $n=5$, then $7\mid x^5-y^5$ implies $7\mid x-y$, but $7^5>1001$. (Also works with $13$ instead of $7$.)
  • $n=6$ is covered by $n=3$.
  • If $n=7$, then $13\mid x^7-y^7$ implies $13\mid x-y$, but $13^7>1001$. (Also works with $7$ or $11$ instead of $13$.)
  • $n=8$ is covered by $n=4$.
  • $n=9$ is covered by $n=3$.
  • $n<10$, because $x^n-y^n\geq(y+1)^n-y^n\geq2^n-1>1001$ if $n\geq10$.

Note that some cases permit alternative approaches, for example $n=7$:
$7\mid x^7-y^7$ implies $7\mid x-y$. Using Newton's binomial theorem this is seen to imply $7^2\mid x^7-y^7$, but $7^2\nmid1001$.
Similarly for $n=11$ or $n=13$.

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  • $\begingroup$ why $n=9$ is covered by $n=3?$. Yes, an solution for $n=9$ can be reduced to an solution for $n=3$ but the reduction is not bijection. $\endgroup$
    – Leox
    May 23, 2015 at 13:19
  • $\begingroup$ Once we've shown that $x^3-y^3=1001$ does not have any solution, we certainly known that $(x^3)^3-(y^3)^3=1001$ does neither. Indeed the reduction is not a bijection and that's why for example we can't conclude the case $n=3$ from $n=9$. But conversely, we can. $\endgroup$ May 23, 2015 at 14:36
  • $\begingroup$ What about the case $x-y=1?$ $\endgroup$
    – Leox
    May 23, 2015 at 14:55
  • $\begingroup$ Please be more specific what your doubts are. Note that every solution of $a^9-b^9=1001$ corresponds to a solution of $x^3-y^3=1001$ (namely $x=a^3$ and $y=b^3$), but the converse need not be true. $\endgroup$ May 23, 2015 at 14:59

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