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| 1. |
If A=m3×n×p2, where m, n and p are prime numbers, which of the following is a perfect square? |
| Answer» <html><body><p><strong>Answer:</strong></p><p>More generally, we show that the only integers n for which 1+n+n2+n3+n4 is a square are −1 , 0 , and 3 .</p><p></p><p>Let <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>(n)=4n4+4n3+4n2+4n+4 , n∈Z . From</p><p></p><p>f(n)=(2n2+n)<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>+3(n+23)2+83 ,</p><p></p><p>we have f(n)>(2n2+n)2 .</p><p></p><p>From</p><p></p><p>f(n)=(2n2+n+1)2−(n+1)(n−3)…(1) ,</p><p></p><p>we have f(n)<(2n2+n+1)2 <a href="https://interviewquestions.tuteehub.com/tag/except-447107" style="font-weight:bold;" target="_blank" title="Click to know more about EXCEPT">EXCEPT</a> when −1≤n≤3 .</p><p></p><p>Since f(n) lies between (2n2+n)2 and (2n2+n+1)2 except when n∈{−1,0,1,2,3} , f(n) is a square implies n∈{−1,0,1,2,3} .</p><p></p><p>From eqn. (1) , f(−1) and f(3) are squares, as is f(0)=4 , whereas f(1)=42+4=20 and f(2)=112+3=124 are not.</p><p></p><p>Since f(n) is even, f(n) is a square if and only if 14f(n) is a square. <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> we have shown that for n∈Z </p><p></p><p>1+n+n2+n3+n4 is a square ⇔n∈{−1,0,3} . ■</p></body></html> | |