\begin{cases} x + y = 4 \\ x - y = 0 \end{cases}

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Question: Can the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 }[/tex] ( i.e. [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ) helps us find [tex]\triangle n(\lambda) = 0[/tex] in the interval, [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] ?

Remark: We are seeking any formulation that yields a global minimum over [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex].

Question: What is the power series expansion of [tex]\sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex] about [tex]\lambda_{0 } = \left \lfloor{\frac{\sqrt{I}}{4}} \right \rfloor[/tex] ?

Remarks: Our questions (above) are redundant, and they have dead-end answers!

Tentative Remark: Yes! The power series expansion is not appropriate here. But the Fourier series expansion is appropriate here.

[tex]\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex].

Moreover, [tex]\triangle n(I, \lambda) = \frac{1}{2} - \frac{1}{\pi} * \sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k}[/tex]

where [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] for a given

I such that

I = p * q.

Tentative Remark: If our explicit definition of [tex]\triangle n(I, \lambda)[/tex] holds, then we have almost solved our integer factorization problem!

Question: What is the solution of [tex]\triangle n(I, \lambda) = 0[/tex] ?