# Find the next least number N that is N+1 = X^2 and (N/2)+1 = y^2?

To go with this following task to create a proper equation ?

Many numbers, especially from smaller ones, can rightly claim that they are of particular interest to scientists. In the kingdom of the right squares, the number 48 is particularly curious. If you add to it 1, you get a square [48 + 1 = 49 = 7x7], and if you split it half and add 1 to the result, you will get another exact square [(48/2 = 24) + 1 = 25 = 5x5 ]. Both conditions are quite trivial, but together, they are far less satisfied.

In fact, 48 is the smallest number that satisfies both conditions. Can you find the next least number that meets both conditions?

• The numbers can easier be found by searching for positive integers $n$ such that both $n$ and $2n-1$ are perfect squares. This boils down to find the positive integers $k$ such that $2k^2-1$ is a perfect square. This leads to a variant of the Pell-equation : $$a^2-2b^2=-1$$ – Peter Jan 10 at 9:36
• The first $4$ numbers satisfying the conditions are : $$48\ 1680\ 57120\ 1940448$$ – Peter Jan 10 at 9:41
• @Peter why it is -1 not 0 ? – Adelin Jan 10 at 13:25
• The square $m^2$ must be of the form $2k^2-1$, leading to $$m^2=2k^2-1$$ which is equivalent to $$m^2-2k^2=-1$$ or in other letters $$a^2-2b^2=-1$$ – Peter Jan 10 at 13:27
• Also note that $(n/2+1)\cdot 2-1=n+1$ – Peter Jan 10 at 13:28