# Polynomials with integer coefficients

Through definitions, theorems and my professor the following is true:

• The product of any two odd integers is odd.
• The sum and difference of any two odd integers are even.
• The sum, product and difference of any two even integers are even.

When expression of the form $(x-r)(x-s)$ are multiplied out a quadratic polynomial is obtained. For example $(x-2)(x+7)$ = $x^2 + 5x -14$

But I cannot explain how $x^2 – 1253x + 255$ can’t be written as a product of two polynomials with integer coefficients.

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You are looking for two integers, the product of which is $255$ and the sum of which is $-1253$. Can the absolute value of the product be so much smaller than the absolute value of the sum? – Todd Wilcox Feb 21 '13 at 20:59
The only way to factor a quadratic is $(x-r)(x-s) = x^2 - (r+s)x + rs$, so in your case $r+s=1253$ and $rs = 255$. That last one leaves very few choices for integer $r,s$, none of which will satisfy the first criterion... – gt6989b Feb 21 '13 at 20:59

Suppose there are integers $r,s$ such that $rs=255, \ r+s=1253$.

From $rs=255$ we get that both $r$ and $s$ are odd and from $r+s=1253$ a contradiction.

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Given the introduction aboutparities, this is apparently the solution that was intended. – Hagen von Eitzen Feb 21 '13 at 21:31
Yes thank you, This is the perfect answer which relates to the parities mentioned above. – CodingWonders90 Feb 21 '13 at 21:58

If you could write $x^2-1253x+255=(x-r)(x-s)$, you would have $rs=255, r+s=1253$ Given that $255=3\cdot 5 \cdot 17$ there are not many choices for $r,s$ and none of them add up to $1253$.

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The key is to use parity, as hinted. $\rm\:mod\ 2\!:\ f(x) \equiv x^2 + x + 1\:$ has no roots, since $\rm\ f(0)\equiv 1\equiv f(1).\:$ Hence there are no integer roots (else reducing mod $\,2\,$ yields a root mod $\,2$). More generally:

Parity Root Test $\$ A polynomial $\rm\:f(x)\:$ with integer coefficients has no integer roots if its constant coefficient and coefficient sum are both odd.

Proof $\$ The test verifies that $\rm\ f(0) \equiv 1\equiv f(1)\ \ (mod\ 2)\:,\$ i.e. that $\rm\:f(x)\:$ has no roots modulo $2$, hence no integer roots. $\$ QED

This test extends to many other rings which have a "sense of parity", i.e. an image $\cong \Bbb Z/2,\:$ for example, various algebraic number rings such as the Gaussian integers.

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There are several ways to solve this problem...

1/ One way is to use the Rational Roots Theorem, which said if I can find a ration root for the given polynomial, that means the polynomial is reducible in the set of rationals (thus in the set of integers). But a simpler way to check is that if there is such a root, say r, exists, then r must divides the "constant" coefficient, which is 255. Now we can list all the possibilities for the divisors, plug them in and try them all out...

2/ Or, we can use the idea in Eric's answer

3/ The long way (and may be impossible), is to use the Eisenstein' criterion. The idea is to come up with a prime p such that it divides all coefficients (except the one of the highest power term), it doesn't divides the highest-power coefficient, and p^2 doesn't divide the last "constant" coefficient. If such prime exists, then we say the polynomial is not reducible (i.e. can't factor) over the field of rationals (thus, over the set of integers)

I said this may be impossible because sometimes there is no such prime exist.

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Well, if $x^2 - 1253x + 255 = (x - a) (x - b)$ for two integers $a$ and $b$, then $ab = 255$ and $a + b = 1253$. The number $255 = 3*5*17$, so there are only a few possibilities:

$a = 1, b = 255$

$a = 3, b = 85$

$a = 15, b = 17$

$a = 85, b = 3$

$a = 255, b = 1$

And clearly none of those satisfy $a + b = 1253$. More generally, there's a theorem called the rational root theorem which tells you when a given polynomial with integer coefficients has a rational/integer root:

http://en.wikipedia.org/wiki/Rational_root_theorem

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Yeah I misread it as 225 at first – Eric Haengel Feb 21 '13 at 21:04

$$x^2 – 1253x + 255=(x-x_1)(x-x_2)$$ where $$x_{1,2}=\frac{1253\pm\sqrt{1253^2-4\cdot255}}{2}=\frac{1253\pm 1252.59}{2}\notin\mathbb Z$$

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