# Riddle: which is the next bigger number? [closed]

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

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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 $x=41^2-1 = 1680$.
The smallest i could get was $1680$. ($1681=41^2, 841=29^2$).