document.write( "Question 1208953: Suppose a >= 2 and n is a natural number larger than 1.
\n" ); document.write( "How can I prove that if n is odd, then a^n+1 is not prime? \n" ); document.write( "
Algebra.Com's Answer #847495 by math_tutor2020(3816)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "It wasn't stated, but I'm assuming 'a' is an integer.
\n" ); document.write( "I'll also assume the +1 is not in the exponent.
\n" ); document.write( "If the +1 was in the exponent, then it's very easy to show that a^(n+1) is always composite regardless if n was even or odd.\r
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\n" ); document.write( "\n" ); document.write( "To factor a polynomial, it's often most efficient to look for the roots.
\n" ); document.write( "For instance, to factor x^2+5x+6, set it equal to zero and use the quadratic formula. You should determine the roots are x = -2 and x = -3.
\n" ); document.write( "Those lead to the factors (x+2) and (x+3). Hence x^2+5x+6 = (x+2)(x+3).
\n" ); document.write( "You can use guess-and-check for small values like this, but it gets tricky when trying to factor something like x^2+31x+238.\r
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\n" ); document.write( "\n" ); document.write( "Anyways, the roots are closely connected to the factorization of the polynomial.
\n" ); document.write( "Consider the polynomial equation y = (x^n)+1.
\n" ); document.write( "Using the rational root theorem, we see that x = -1 is a potential root.
\n" ); document.write( "If x = -1 and n is odd, then x^n+1 = (-1)^(2k+1)+1 = -1*( (-1)^2 )^k+1 = -1+1 = 0 which confirms that x = -1 is a root when n is odd.
\n" ); document.write( "x = -1 being a root leads to x+1 being a factor after adding 1 to both sides.\r
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\n" ); document.write( "\n" ); document.write( "Use a graphing tool like GeoGebra or Desmos to look at the graphs of y = (x^3)+1, y = (x^5)+1, y = (x^7)+1, etc.
\n" ); document.write( "All of those graphs have x = -1 as an x intercept and hence (x+1) as a factor.
\n" ); document.write( "This will prove (x^n)+1 is composite when n is odd and x is an integer.
\n" ); document.write( "By extension it does the same for (a^n)+1.
\n" ); document.write( "Here are a few select examples:
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document.write( "n = 3:\r\n" );
document.write( "	a = 2;  (a^n)+1 = (2^3)+1 = 9 is composite because 9 = 3*3\r\n" );
document.write( "	a = 3;  (a^n)+1 = (3^3)+1 = 28 is composite because 28 = 4*7\r\n" );
document.write( "	a = 4;  (a^n)+1 = (4^3)+1 = 65 is composite because 65 = 5*13\r\n" );
document.write( "n = 5:\r\n" );
document.write( "	a = 2;  (a^n)+1 = (2^5)+1 = 33 is composite because 33 = 3*11\r\n" );
document.write( "	a = 3;  (a^n)+1 = (3^5)+1 = 244 is composite because 244 = 4*61\r\n" );
document.write( "	a = 4;  (a^n)+1 = (4^5)+1 = 1025 is composite because 1025 = 5*205\r\n" );
document.write( "n = 7:\r\n" );
document.write( "	a = 2;  (a^n)+1 = (2^7)+1 = 129 is composite because 129 = 3*43\r\n" );
document.write( "	a = 3;  (a^n)+1 = (3^7)+1 = 2188 is composite because 2188 = 4*547\r\n" );
document.write( "	a = 4;  (a^n)+1 = (4^7)+1 = 16385 is composite because 16385 = 5*3277\r\n" );
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\n" ); document.write( "Note that
  • When a = 2, a+1 = 3 is a factor of (a^n)+1.
  • When a = 3, a+1 = 4 is a factor of (a^n)+1.
  • When a = 4, a+1 = 5 is a factor of (a^n)+1.
  • And so on.
All of what is mentioned is where n is odd. This wraps up the proof. \r
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\n" ); document.write( "\n" ); document.write( "If you're curious what happens when n is even then the graph of y = (x^n)+1 is always above the x axis.
\n" ); document.write( "It won't have any real number roots.
\n" ); document.write( "The roots will be complex numbers of the form a+bi where i = sqrt(-1)
\n" ); document.write( "An example would be y = x^2+49 which has the complex roots 7i and -7i.
\n" ); document.write( "So it's not clear if (a^n)+1 is composite or prime when n is even.
\n" ); document.write( "Turns out it depends. Here are some examples.
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document.write( "n = 2:\r\n" );
document.write( "	a = 2;  (a^n)+1 = (2^2)+1 = 5 is prime (the only factors are 1 and 5)\r\n" );
document.write( "	a = 3;  (a^n)+1 = (3^2)+1 = 10 is composite since 2*5 = 10\r\n" );
document.write( "	a = 4;  (a^n)+1 = (4^2)+1 = 17 is prime (the only factors are 1 and 17)\r\n" );
document.write( "n = 4:\r\n" );
document.write( "	a = 2;  (a^n)+1 = (2^4)+1 = 17 is prime (the only factors are 1 and 17)\r\n" );
document.write( "	a = 3;  (a^n)+1 = (3^4)+1 = 82 is composite since 82 = 2*41\r\n" );
document.write( "	a = 4;  (a^n)+1 = (4^4)+1 = 257 is prime (the only factors are 1 and 257; see \"primality check\" below)\r\n" );
document.write( "n = 6:\r\n" );
document.write( "	a = 2;  (a^n)+1 = (2^6)+1 = 65 = 5*13 is composite\r\n" );
document.write( "	a = 3;  (a^n)+1 = (3^6)+1 = 730 = 10*73 is composite\r\n" );
document.write( "	a = 4;  (a^n)+1 = (4^6)+1 = 4097 = 17*241 is composite\r\n" );
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Primality Check: To determine if 257 is prime or not, divide it over each item in the list of primes smaller than sqrt(257) = 16.0312
You'll divide 257 over the following primes: 2,3,5,7,11, and 13
You should find that the result of each division is a non-integer decimal value.
This proves none of those values are a factor of 257 and proves 257 is prime.

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