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I am looking at this problem and I am confused on how one could easily compute this. Is it just the intersection of either non-quadratic residue or quadratic residues of the respective p and q?

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What is the question again? –  Jyrki Lahtonen Mar 26 '12 at 19:51

3 Answers 3

up vote 1 down vote accepted

Firstly the case $i=j=1$ is simple since the number $1$ will do this.

Secondly when $i = -1$ and $j=1$ we can use the number $-1$ since this is a QR mod $p'$ (prime) if and only if $p'\equiv 1$ mod $4$.

Thirdly when $i=j=-1$ we can use the number $2$ since this is a QR mod $p'$ if and only if $p\equiv 1,7$ mod $8$.

For the final case we may use $-2$ since we can see that the product of the $i,j$ values in the previous two cases match what we want (essentially we are using the homomorphism property of Legendre symbols here).

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Hi fretty, why is mod 8 or mod 4 used or derived from? Thanks –  user27617 Mar 26 '12 at 15:09
    
These are the simple rules about when $-1$ or $2$ are QR's mod $p$ –  fretty Mar 26 '12 at 18:09
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Sorry. I have problems parsing this. What are the variables $i$ and $j$? Where did they come from? –  Jyrki Lahtonen Mar 26 '12 at 19:50
    
The question seems to have been edited! –  fretty Mar 26 '12 at 21:12

If you want to find $x$ such that it is a QR mod p and a NR mod q for example, you can just find a QR mod p (say m), find a NR mod q(say n), then use Chinese remainder theorem to solve

$$\begin{cases}x \equiv m & (\operatorname{mod} p) \\ x \equiv n & (\operatorname{mod} q) \end{cases}$$

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Hi user27126, could you please provide an example for say i=1 and j=-1. I am not sure how the CR theorem can help here. Thank you. –  user27617 Mar 26 '12 at 13:59
    
Let's take $m = 0$, $n = 2$. You want to find x = 0 (mod 19), x = 2 (mod 37). Just use chinese remainder theorem to solve this congruence. –  user27126 Mar 26 '12 at 17:08

The solution by user27126 is fine, but you can cut a few corners. For example, there is one kind of number $n$ that is guaranteed to be a quadratic residue mod $p$ for all $p$ not dividing $n$ - do you know what I'm referring to? Also, you should know when $-1$ is a residue, that should settle another case without any need to resort to Chinese remainders. If you know when 2 is a residue, there goes another case. And so on.

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