-2
$\begingroup$

How to solve this puzzle? Follow this link

$\endgroup$

closed as off-topic by Hans Lundmark, Claude Leibovici, daw, JonMark Perry, Davide Giraudo Oct 29 '15 at 10:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hans Lundmark, Claude Leibovici, daw, JonMark Perry, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

The subsequent smallest value of $x$ is $1680$.

We need $x$ such that $$x+1 = m^2 \text{ and }x/2+1 = n^2$$ Eliminating $x$, we need $m$ and $n$ such that $$m^2 + 1 = 2n^2 \text{ or }m^2-2n^2=-1$$ This is an example of Pell's equation, see here and here. The approach is to guess the smallest positive solution, which in this case is $(n_1,m_1)=(1,1)$. All the remaining solutions are given by the Brahmagupta's identity, i.e., $$(m_{k+1},n_{k+1}) = (3m_k+4n_k,2m_k+3n_k)$$ This gives us that $(m_2,n_2) = (7,5)$, i.e., $x=7^2-1 = 48$.

Next, we obtain $(m_3,n_3) = (3\cdot7+4\cdot5,2\cdot7+3\cdot5) = (41,29)$, which gives us that $x=41^2-1 = 1680$.

$\endgroup$
1
$\begingroup$

The smallest i could get was $1680$. ($1681=41^2, 841=29^2$).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.