# Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to calcu

• Jun 7th 2012, 07:31 AM
math2011
Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to calcu
Given that $907$ and $997$ are primes, use the fact that $907 \equiv -997 \pmod{28}$ to calculate $\left( \frac{7}{997} \right) - \left( \frac{7}{907} \right)$.

I am a bit confused about the question to start with, is $\left( \frac{7}{997} \right) - \left( \frac{7}{907} \right)$ about fractions or quadratic residues?

How can I approach this question?
• Jun 7th 2012, 10:27 PM
Sarasij
Re: Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to c
Quote:

Originally Posted by math2011
Given that $907$ and $997$ are primes, use the fact that $907 \equiv -997 \pmod{28}$ to calculate $\left( \frac{7}{997} \right) - \left( \frac{7}{907} \right)$.

I am a bit confused about the question to start with, is $\left( \frac{7}{997} \right) - \left( \frac{7}{907} \right)$ about fractions or quadratic residues?

How can I approach this question?

Its not understandable to me at all.
• Jun 8th 2012, 04:35 AM
ILikeSerena
Re: Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to c
Hi math2011! :)

Those will be quadratic residues if only because otherwise the parentheses would be redundant.
Furthermore, it's not an interesting problem if they were only regular fractions.

The fact that the difference is requested, suggest that from each possible quadratic residue of the first, you can draw a conclusion about the quadratic residue of the second.

So if there is an x such that $7 \equiv x^2 \pmod{907}$ it should follow that:
- either you can also find an y with: $7 \equiv y^2 \pmod{997}$
- or you can proof that the congruence with y does not hold for any y.

And if there is no such x, the inverse should hold.
Your result would then be either -2, 0, or 2.

To be honest, this is as far as I got.
• Jun 9th 2012, 07:12 AM
math2011
Re: Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to c
Thank you for helping me understand the question.
• Jun 9th 2012, 07:26 AM
ILikeSerena
Re: Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to c
You're welcome. :)

Actually I was hoping for someone else to respond, but no one did. :(
• Jun 9th 2012, 07:45 AM
ILikeSerena
Re: Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to c
I can tell you that the result is 0, because:

$7^{(907-1)/2} \equiv -1 \pmod{907}$

and

$7^{(997-1)/2} \equiv -1 \pmod{997}$

according to wolframalpha.com.

That is:

$\left( \frac 7 {907} \right) = -1$

and also

$\left( \frac 7 {997} \right) = -1$.
• Jun 25th 2012, 01:03 AM
math2011
Re: Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to c
I missed something - there is actually a Lemma in my lecture notes for this.

For $a \in \mathbb{N}$ and primes $p,q$ with $p \equiv \pm q \pmod{4a}$, $\left(\frac{a}{p}\right) = \left(\frac{a}{q}\right)$.

It is straightforward to solve this problem using this Lemma.

The proof of the Lemma is said to be difficult.
• Jun 25th 2012, 12:28 PM
ILikeSerena
Re: Given that 907 and 997 are primes, use the fact that 907 equiv -997 (mod 28) to c
Ah, okay. That would make it easier. :)