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<h2>Prime square remainders</h2><div class="info" style="cursor:help;width:200px;margin-bottom:10px;"><h3>Problem 123</h3><span style="width:300px;color:#666;">Published on Friday, 16th June 2006, 06:00 pm; Solved by 5862</span></div>
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<p>Let <i>p</i><sub>n</sub> be the <i>n</i>th prime: 2, 3, 5, 7, 11, ..., and let <i>r</i> be the remainder when (<i>p</i><sub>n</sub><img src='images/symbol_minus.gif' width='9' height='3' alt='&minus;' border='0' style='vertical-align:middle;' />1)<sup><i>n</i></sup> + (<i>p</i><sub>n</sub>+1)<sup><i>n</i></sup> is divided by <i>p</i><sub>n</sub><sup>2</sup>.</p>
<p>For example, when <i>n</i> = 3, <i>p</i><sub>3</sub> = 5, and 4<sup>3</sup> + 6<sup>3</sup> = 280 <img src='images/symbol_cong.gif' width='9' height='11' alt='&equiv;' border='0' style='vertical-align:middle;' /> 5 mod 25.</p>
<p>The least value of <i>n</i> for which the remainder first exceeds 10<sup>9</sup> is 7037.</p>
<p>Find the least value of <i>n</i> for which the remainder first exceeds 10<sup>10</sup>.</p>
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