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This is very easy to prove using modulo arithmetic. Basically, what you do is to look at the remainder when a number (n) is divided by 3. Let k(mod 3) represent the remainder when a number is divided by 3.
Since the divisor is 3, there are only 3 possible values for k, that is:
n = 0(mod 3), 1(mod3) or 2(mod3).
Suppose n = 0(mod 3)
then n2 + 1 = 0 + 1(mod 3) = 1(mod 3)
so that n2 + 1 leaves a remainder of 1 when divided by 3 and so is not divisible by 3.
Suppose n = 1(mod 3)
then n2 + 1 = 12 + 1(mod 3) = 2(mod 3)
so that n2 + 1 leaves a remainder of 2 when divided by 3 and so is not divisible by 3.
Suppose n = 2(mod 3)
then n2 + 1 = 22 + 1(mod 3) = 5(mod 3) = 2(mod 3)
so that n2 + 1 leaves a remainder of 2 when divided by 3 and so is not divisible by 3.
Thus, for all possible values of n, division by 3 leaves a positive remainder. And so the result follows.
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104
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4
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104
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What is the product of four x squared minus one over two x squared minus five x minus three and x squared minus six x plus nine over two x squared plus five x minus three?
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