<PRE><B>Question 192325</B>
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I can&#39;t do that.  Nobody can.  That&#39;s because, in general, *[tex \LARGE  a^2 + b^2] &lt;i&gt;&lt;b&gt;can&lt;/b&gt;&lt;/i&gt; be factored -- just not over the real numbers.  That is to say, real number expressions &lt;i&gt;&lt;b&gt;p&lt;/b&gt;&lt;/i&gt; and &lt;i&gt;&lt;b&gt;q&lt;/b&gt;&lt;/i&gt; do not exist such that *[tex \LARGE  pq = a^2 + b^2].  However, if you consider the complex numbers of the form *[tex \LARGE  \alpha + \beta i] where *[tex \LARGE  i] is defined as *[tex \LARGE  i^2 = -1], then *[tex \LARGE  a^2 + b^2] factors quite nicely to *[tex \LARGE  (a + bi)(a - bi)].



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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