document.write( "Question 794840: Determine all integers n such that n^4-4n^3+15n^2-30n+27 is a prime number \n" ); document.write( "

Algebra.Com's Answer #480818 by KMST(5328) $\"\"$ $\"About$

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$\"n%5E4-4n%5E3%2B15n%5E2-30n%2B27%3E0\"$ for all real values of $\"n\"$ ,
\n" ); document.write( "so there are no factors of the form $\"%28x-a%29\"$ ,
\n" ); document.write( "but I can still factor that polynomial into the product of two quadratic polynomials:
\n" ); document.write( " $\"n%5E4-4n%5E3%2B15n%5E2-30n%2B27=%28n%5E2%2Bbn%2Bc%29%2A%28n%5E2%2Bdn%2Be%29\"$
\n" ); document.write( "Multiplying, I get
\n" ); document.write( "

\n" ); document.write( "I need to find b, c, d, and e such that
\n" ); document.write( " $\"ce=27\"$ , so the set $\"c%2Ce\"$ is {1,27}, or {-1,-27}, or {3,9}, or {-3,-9}
\n" ); document.write( " $\"b%2Bd=-4\"$
\n" ); document.write( " $\"c%2Bbd%2Be=15\"$ and
\n" ); document.write( " $\"eb%2Bcd=-30\"$
\n" ); document.write( "Since c and e are both odd, $\"c%2Be\"$ will be even,
\n" ); document.write( "and $\"bd\"$ must be odd to make $\"c%2Bbd%2Be\"$ odd, let alone $\"c%2Bbd%2Be=15\"$
\n" ); document.write( "To make $\"bd\"$ odd, b and d must both be odd.
\n" ); document.write( "They have to add up to $\"-4\"$ too to make $\"b%2Bd=-4\"$
\n" ); document.write( "There are many options, but With the set {b,d}={-1,-3} I could get
\n" ); document.write( " $\"b%2Bd=-1-3=-4\"$ and
\n" ); document.write( " $\"bd=3\"$ , which with {c,e}={3,9},
\n" ); document.write( "added to $\"c%2Be=3%2B9\"$ to make $\"c%2Bbd%2Be=3%2B9%2B3=15\"$
\n" ); document.write( "Trying the options left for the order of the numbers in each set,
\n" ); document.write( " $\"eb%2Bcd=-30\"$ could be made pairing them as $\"9%28-3%29%2B3%28-1%29=+-27-3=-30\"$
\n" ); document.write( "with $\"e=9\"$ , $\"b=-3\"$ , $\"c=3\"$ , and $\"d=-1\"$ .
\n" ); document.write( "So $\"n%5E4-4n%5E3%2B15n%5E2-30n%2B27=%28n%5E2-3n%2B3%29%2A%28n%5E2-n%2B9%29\"$
\n" ); document.write( "With $\"n\"$ being an integer, the factors
\n" ); document.write( " $\"%28n%5E2-3n%2B3%29\"$ and $\"%28n%5E2-n%2B9%29\"$ are integers.
\n" ); document.write( "For their product to be a prime number, one factor must be 1,
\n" ); document.write( "and the other factor must be a prime number.
\n" ); document.write( "
\n" ); document.write( "Making the first factor equal to 1, we get
\n" ); document.write( " $\"n%5E2-3n%2B3=1\"$ --> $\"n%5E2-3n%2B3-1=0\"$ --> $\"n%5E2-3n%2B2=0\"$ --> $\"%28n-2%29%28n-1%29=0\"$ --> $\"system%28n=2%2C%22or%22%2Cn=1%29\"$
\n" ); document.write( " $\"n=2\"$ --> $\"n%5E2-n%2B9=2%5E2-2%2B9=4-2%2B9=11\"$ leads us to
\n" ); document.write( "the solution $\"highlight%28n=2%29\"$ ,
\n" ); document.write( "which makes $\"n%5E4-4n%5E3%2B15n%5E2-30n%2B27=%28n%5E2-3n%2B3%29%2A%28n%5E2-n%2B9%29=1%2A11=11\"$
\n" ); document.write( " $\"n=1\"$ also makes $\"n%5E2-3n%2B3=1\"$ , but it makes
\n" ); document.write( " $\"n%5E2-n%2B9=1%5E2-1%2B9=1-1%2B9=9\"$ and 9 is not a prime number.
\n" ); document.write( "
\n" ); document.write( "Making the second factor equal to 1, we get
\n" ); document.write( " $\"n%5E2-n%2B9=1\"$ --> $\"n%5E2-n%2B9-1=0\"$ --> $\"n%5E2-n%2B8=0\"$ , which has no real solutions.
\n" ); document.write( "
\n" ); document.write( "NOTE: When you find a problem that seems to difficult for mere humans, like us, try the artofproblemsolving community forum. The members there can't resist a challenge, and if it proves too difficult for the level where you post it, the moderators may bump it up. Start where you think is to low (try high school intermediate or lower) so you do not offend them by posting problems that are insultingly easy for them. I bet some 12 or 13-year old will show you how easy your problem is. \n" ); document.write( "

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