document.write( "Question 1151635: Solve the equation for the value of n: \"x%5E%28n-1%29%2Bx%5E%28n%2B1%29=320\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E%28n-1%29%2Bx%5E%28n%2B1%29=320\"
\n" ); document.write( "\"x%5E%28n-1%29%2Bx%5E%28n%2B1%29=256%2B64\"
\n" ); document.write( "\"x%5E%28n-1%29%2Bx%5E%28n%2B1%29=2%5E6%2B2%5E8\"
\n" ); document.write( "So, \"2%5E%28n-1%29=2%5E6\"
\n" ); document.write( "\"n-1=6\"
\n" ); document.write( "\"n=7\"\r
\n" ); document.write( "\n" ); document.write( "Can I solve the question this way ? If not , please solve it in the right way . Thanks
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Algebra.Com's Answer #773449 by ikleyn(52811)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Let start discussing the problem's formulation.\r\n" );
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document.write( "If you allow x and \"n\" to be any real numbers, then, probably, other solutions (non-integer, rational or irrational) may exist.\r\n" );
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document.write( "But, probably, the problem restricts you by integer values of x and \"n\" --- it would be very natural.\r\n" );
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document.write( "So, let's assume that we are looking for integer positive x and \"n\", only.\r\n" );
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document.write( "Then your approach is only PARTLY good : you just found one solution, but you did not guarantee that there is no other solution.\r\n" );
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document.write( "In your post, you even did not touch this issue.\r\n" );
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document.write( "In this sense, your solution is not complete.\r\n" );
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document.write( "There is another approach.\r\n" );
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document.write( "The starting equation can be transformed into other, equivalent equation\r\n" );
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document.write( "    \"x%5E%28n-1%29%2A%281%2Bx%5E2%29\" = 320\r\n" );
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document.write( "    \"x%5E%28n-1%29%2A%281%2Bx%5E2%29\" = \"2%5E6%2A5\".\r\n" );
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document.write( "Having it in this form, it is easy to prove that the only solution is\r\n" );
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document.write( "    x = 2, n = 7,\r\n" );
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document.write( "using the divisibility properties of integer numbers.\r\n" );
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\n" ); document.write( "\n" ); document.write( "At this point I'd like to complete my explanation.\r
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