Bound on the number of solutions of a specific Diophantine equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:46:18Z http://mathoverflow.net/feeds/question/35375 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35375/bound-on-the-number-of-solutions-of-a-specific-diophantine-equation Bound on the number of solutions of a specific Diophantine equation N. Kumar 2010-08-12T17:43:22Z 2010-08-12T23:37:04Z <p>Falco had asked a question regarding sum equals to product ( <a href="http://mathoverflow.net/questions/35150/sum-equals-product" rel="nofollow">http://mathoverflow.net/questions/35150/sum-equals-product</a>)</p> <p>I have a question in the orthogonal direction. Suppose $X_1,X_2,...,X_n$ are variables and we allow $X_i$'s to take only natural numbers. Look at the following Diophantine equation $X_1+X_2+ \dots + X_n = X_1 X_2 \ldots X_n$. Any solution of this equation satiesfies the property that the sum of the entries is equal to their product.</p> <p>It is easy to see that for every $n$, there are only finitely many solutions of the above equation, denote that number by $f(n)$. It is easy to see that there is no absolute constant $k \in \mathbb{N}$ such that $f(n) &lt; k$ for every $n$. (look at the sequence $x_n= n!+1$, then $f(x_n) > n$, for $n \geq 5$)</p> <p>If $(x_1,..., x_n)$ is a solution of the above equation then we have $\prod_{i=1}^{n-1} x_i &lt; n$. From here one can have a very crude bound for $f(n)$.</p> <p>Question: 1) What is the best upper bound for $f(n)$? 2) Is there an asymptotic behaviour of $f(n)$ as $n$ tends to infinity.</p> http://mathoverflow.net/questions/35375/bound-on-the-number-of-solutions-of-a-specific-diophantine-equation/35404#35404 Answer by Gerry Myerson for Bound on the number of solutions of a specific Diophantine equation Gerry Myerson 2010-08-12T23:37:04Z 2010-08-12T23:37:04Z <p>D24 in Guy's Unsolved Problems In Number Theory: For $k>2$ the equation $$a_1a_2\cdots a_k=a_1+a_2+\cdots+a_k$$ has the solution $a_1=2$, $a_2=k$, $a_3=a_4=\cdots=a_k=1$. Schinzel showed that there is no other solution in positive integers for $k=6$ or $k=24$. Misiurewicz has shown that $k=2,3,4,6,24,114,174$ and 444 are the only $k&lt;1000$ for which there is exactly one solution. The search has been extended by Singmaster, Bennett and Dunn to $k\le1440000$. They let $N(k)$ be the number of different 'sum = product' sequences of size $k$, and conjecture that $N(k)>1$ for all $k>444$. They find that $N(k)=2$ for 49 values of $k$ up to 120000, the largest being 6174 and 6324, and conjecture that $N(k)>2$ for $N>6324$. They also find that $N(k)=3$ for 78 values of $k$ in the same range, the largest being 7220 and 11874, and conjecture that $N(k)>3$ for $k>11874$; also that $N(k)\to\infty$. </p> <p>Guy gives many references. </p>