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Diophantine equations are polynomial equations $F=0$, or systems of polynomial equations $F_1=\ldots=F_k=0$, where $F,F_1,\ldots,F_k$ are polynomials in either $\mathbb{Z}[X_1,\ldots,X_n]$ of $\mathbb{Q}[X_1,\ldots,X_n]$ of which it is asked to find solutions over $\mathbb{Z}$ or $\mathbb{Q}$. Topics: Pell equations, quadratic forms, elliptic curves, abelian varieties, hyperelliptic curves, Thue equations, normic forms, K3 surfaces ...
3
votes
Accepted
Triangular repdigits
Following user523984's suggestion in the comments:
From triangle = repdigit $$\frac{k(k+1)}2 = \frac{d(10^j-1)}9$$ we get $$k = \frac{-1 \pm \frac13 \sqrt{9 + 8d(10^j-1)}}{2}$$ so we require $9 + 8d(1 …
1
vote
Accepted
Only trivial solution to a pair of constrained linear diophantine equations
The answer to the first question is negative.
Let $A$ denote the set of weights $\{a_i\}$. Strengthen the first constraint to $0 \le x_i \le 2$.
If we have two different subsets $S_1, S_2 \subset A$ o …
9
votes
Solving functional equation $f(xy)=f(x+y)$ and Diophantine equations
$7 \sim 12$ via $3, 4$
$12 \sim 35$ via $5, 7$
$35 \sim 264$ via $11, 24$
$264 \sim 41$ via $8, 33$
$41 \sim 420$ via $20, 21$
$420 \sim 43$ via $15, 28$
$43 \sim 156$ via $4, 39$
$156 \sim 25$ via $ …
4
votes
Non-trivial solution to $\sum^{n}_{i=1}\sum^{n}_{j=1,j\ne i}(x_{i})^{(x_j)}=(\sum^{n}_{i=1}x...
Split $x_i$ into $z$ zeroes and a partition of $n$ into $k$ (non-zero) parts, $\lambda_j$. Then your equality can be rewritten as $$z(z-1) + kz + \mathop{\sum\sum}_{i \neq j} \lambda_i^{\lambda_j} = n …