How should I prove this question?

Prove :

If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8.

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- September 12th 2006, 11:25 PMbeta12square polynomials (mod 8 )
How should I prove this question?

Prove :

If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8. - September 13th 2006, 08:36 AMThePerfectHacker
It seems to me it is a direct connection to quadradic residues.

Remember from field theory that a polynomial of degree 2 or 3 is reducible over a polynomial ring (in this case the integers under multiplication modulo 8) if and only if it has a zero.

Thus, what we are saying is that

x^2+ax+1≡0(mod 8)

For some x.

We will express this in a more efficient way,

x^2+ax+1≡0(mod 2^3)

That means the discriminant of this quadradic is a quadradic residue of 2. - September 14th 2006, 12:45 AMbeta12
Hi perfecthacker,

This question wants us to find the possible values of a. i.e. for which non-negative a less than 8 does f(x) factor?

Do you have any idea about that? - September 14th 2006, 01:25 AMmalaygoel
- September 14th 2006, 05:00 AMThePerfectHacker
- September 14th 2006, 05:25 AMbeta12
Let me write the question again clearly.

Prove :

If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8.

What are the possible values of a ? That is , for which non-negative a less than 8 does f(x) factor?

:( still can't figure out the values of a! - September 14th 2006, 05:31 AMThePerfectHacker
It was clearly the first time I was lazy to answer.

---

Since,

x^2+ax+1≡(x+c)^2=x^2+2cx+c^2 (mod 8)

By definition of polynomial equality we need that,

a=2c

1=c^2

.

So for what values of Z_8 do you have that, a number is is its own inverse. Basic computation for 0,1,...,7 shows that: 3,5,7.

Then,

a=2(3)=6

a=2(5)=2

a=2(7)=6

So the two polynomials you have are,

x^2+2x+1

x^2+6x+1 - September 14th 2006, 09:07 AMbeta12
Hi perfecthacker,

I got it now.

Do you know how to prove this question too?

Prove: If a is any integer and the polynomial f(x) = x^2 + ax +1 factors (polymod 9), then there are three distinct non-negative integers y less than 9 such that f(y)≡ 0 ( mod 9). - September 14th 2006, 10:05 AMThePerfectHacker
It factos,

x^2+ax+1 = (x+b)(x+c)

Thus,

x^2+ax+1=x^2+(b+c)x+bc

Thus,

a=b+c

1=bc

Thus, you need to find all values of 'a'

such that you can find integers 1<=b,c<=9

That make the above statement true.

There should be 3 values of 'b' and 'c'.

The first condition: a=b+c is not important.

All you need to find is distinct 'b' and 'c' such that they are inverses (bc=1) and then declare a=b+c. - September 16th 2006, 01:28 AMbeta12
Hi perfecthacker,

Thank you very much.