Smallest quadratic non residue. de; Date: Fri, 09 Jun 2023 11:26: .

Smallest quadratic non residue In ranges to the question of bounding the size of the smallest quadratic non-residue modulo pas a function of p. Our aim in this paper is to establish the distribution of quadratic non-residues in even smaller intervals of size (log p)A with A > 1, for almost all primes p. Soc. 2. The Legendre symbol a p is de ned as a Solving the system (2. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have there are asymptotically as many quadratic residues as non-residues up to p1=4 logc p; and, if all those 1 n < p1=4 p e+ were quadratic residues modulo p, then since a little more than half of the n p1=4 logc p have all their prime factors smaller than p1=4 p e+ , we would have that the quadratic residues up (Deﬁnition) Quadratic Residue: Let pbe an odd prime, a6 0 mod p. , if the congruence (35) has no solution, then q is said to be a quadratic nonresidue (mod p). Our aim in this paper is QUADRATIC RESIDUE ANSD NON-RESIDUES 10. The Burgess bound on the least quadratic non-residue only gives l p ≤ ( 0. Jul 1, 2013 · Let S be an infinite set of nonempty, finite subsets of the positive integers. Therefore, n p is prime. Let g(p, k, b, c) denote the smallest kth power nonresidue in the progression bn + c and let r2(p, b, c) denote the smallest quadratic residue in the progression bn + c. For example, Polya-Vinogradov lets us take N˛ p1/2 logp, giving pθ+o(1) for θ= 1/2. Now, let’s try to see how often n p= 2 (the smallest prime). Apr 25, 2015 · It seems now from the comments that despite the wording of the problem, we are asked to show there are infinitely many irreducibles. IV. Our strategy is then to first bound each q i -th power non-residue individually and then use them to explicitly construct a q 1 , . This is a result of Gauss in the Disquisitiones. By FLT, ap1 1 mod pand p 1 is even. In this case, it is customary to consider 0 as a special case and work within the multiplicative group of nonzero elements of the field (/). A famous result of Gauss 1 Distribution of small quadratic residues and non-residues, Mathematics Seminar, Tezpur University, August 2024. What is the expected size of the k-th smallest quadratic non-residue mod-ulo p? This question is interesting only when we exclude certain obvious choices. Gauss posed the problem of finding the smallest quadratic non-residue $n_p$ modulo a prime p [], and papers over the past century have continued to refine this bound, most recently by Carella [] showing $n_p\ll (\log {p})(\log \log {p})$. We will begin with some background on quadratic non-residues and then give a brief outline of the proof. Go ahead – try it out. Let C(p) be the multiplicative group consisting of the residue classes mod p and let Ck(p) denote the subgroup of the kth powers mod p. Then ais a quadratic residuemod piff ap1 2 1 mod p Proof. Quadratic Residue * Quadratic Non-residue = Quadratic Non-residue. Theorem 6 Second Supplement to the Law of Quadratic Reciprocity Let p be an odd prime. KONYAGIN AND I. In fact, we can even arrange the quadratic residues to be the rst N quadratic residues. Now it is true that an upper bound 2p1=2 on the Sep 13, 2013 · The existing explicit bounds for the least quadratic non-residue and the least prime in an arithmetic progression are improved and the classical conditional bounds of Littlewood for L-functions at s=1 are refined. Introduction Let q>1 be a natural number and let G= (Z/qZ)∗ denote the group of reduced residues (mod q). Can you explain or prove your answer? What about the smallest quadratic non-residue? How big is it, in terms of p? Write a PARI program to test how large the smallest quadratic NON-residue is. everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box Then any a with p - a is a quadratic residue of p i ind ra is even. But what about the smallest non-trivial quadratic residue? That is, the smallest non-square quadratic residue? vs. The best known result is that the least quadratic non-residue is $\ll p^{1/(4\sqrt{e})}$, which follows from Burgess's bound on character sums plus a multiplicative trick of Vinogradov. A non-zero residue a is a quadratic residue (QR) modulo p if x2 ≡a (mod p) has a solution. J. 2 is the lowest non residue possible. 15163 … + o ( 1 ) ) log p / log 2 . 8) we can also conclude that the smallest primitive element, which is also the smallest quadratic non-residue element, of a safe prime p is 11 or higher if and only if it has the form p ≡ 159,199, 279 (mod 280). 42 ≡ 122 Jul 5, 2022 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Jul 20, 2022 · This question is closely related to Linnik's problem on the least quadratic nonresidue for a given prime modulus. (Invited) Gaussian behavior of small quadratic non-residues, Analytic Number Theory and related fields, AMS sion. Cited by (0) 1. $\endgroup$ – Greg Martin Commented Oct 24, 2016 at 4:46 Although Gauss’ celebrated Quadratic Reciprocity Law gives a beautiful criterion to decide whether a given number is a quadratic residue modulo p or not, it is still an open problem to find a small upper bound on the least quadratic non-residue mod p as a function of p, at least when p ≡ 1 (mod 8). Duke Math. Hence any nontrivial bound on |S χ(N)| proves the existence of a quadratic non-residue less than N. Here the evaluation of the least integer n´ among all positive integers n for which ´(n) 6= 0 ;1 is referred as Linnik’s problem. If =p + ku is one of them, p + (lc 8)u is the other. With the aid of Gauss sums it is possible to establish a relation between two important objects in number theory — viz. 117198 … + o ( 1 ) ) log p / log 2 for the smallest Hamming weight l p among prime quadratic non-residues modulo a sufficiently large prime p . We prove unconditionally that for almost all primes p in short intervals, these Applying the theorem to a quadratic character, one has the following corollary. Theorem 5. For example, all small integer multiples of the ﬁrst quadratic non-residue are quadratic non and recall that by the classical bound of Burgess [6] on the smallest quadratic non-residue np we have (1. Assuming the Generalized Riemann Hypothesis for Dirichlet L-functions, Ankeny [1] Preprint submitted to Elsevier November 29, 2014 If there is no quadratic non-residue in [1,N] then S χ(N) = N, so |S χ(N)| attains the trivial upper bound. For example, P olya-Vinogradov lets us take N˛p1=2 logp, giving p +o(1) for = 1=2. We use the Legendre symbol to help keep track of when an integer is a QR. E. Let a6 0 mod p. McGown. If p is an odd prime, let c(p) denote the cardinality of the set {S∈S:S… Subject: efficient determination of smallest quadratic non-residue / t++ works, t=nextprime(t) hangs; From: hermann@stamm-wilbrandt. Thus ais a quadratic residue if and only if ais a root of the polynomial xk 1: This polynomial has at most kroots. Specifically, we improve p is a quadratic residue if there exists an x2Z p such that x2 a(mod p). How do you determine if something is a quadratic residue? If a is an odd number and m = 8, 16, or some higher power of 2, then a is a residue modulo m if and only if a ≡ 1 (mod 8). Modulo an odd prime number p there are (p + 1)/2 residues (including 0) and (p − 1)/2 nonresidues, by Euler's criterion. 2. 16 ) from the quadratic character, that is, from the Legendre symbol is Apr 1, 2013 · The second smallest quadratic non-residue. But the Google Ngram Viewer suggests that most academic the longest sequence of consecutive quadratic residues modulo p. Oct 19, 2022 · We will discuss the Poisson behavior of chains of small quadratic non-residues for almost all primes in short intervals. For every positive integer N there are in nitely many primes p for which 1; 2; :::; N are quadratic residues modulo p, and there ex- The remaining seven chapters explore some interesting applications of the Law of Quadratic Reciprocity, prove some results concerning the distribution and arithmetic structure of quadratic residues and non-residues, provide a detailed proof of Dirichlet’s Class-Number Formula, and discuss the question of whether quadratic residues are Nov 1, 2019 · Given a prime p such that p − 1 has r prime factors q 1,, q r say, we can think of the least primitive root modulo p as the smallest simultaneous q 1,, q r-th power non-residue. We denote the smallest kth power nonresidue in the progression bn + c by g(p,k,b,c), the smallest quadratic residue in the progression bn + c by r2(p,b,c), and the nth smallest prime kth power nonresidue by g„(p,k), n = 0, 1, 2,_ If C(p) is the multiplicative group consisting of the residue classes mod p, then the fcth Nov 19, 2010 · Semantic Scholar extracted view of "On the second smallest prime non-residue" by Kevin J. Introduction Let qbe a natural number and let G= (Z/qZ)∗ denote the group of reduced residues (mod q). 7 (1), he proved that this least quadratic non-residu is O(p*)e for any fixed a > \e~x^. Examples 7. Since/c and/ 8 are incongruent rood 3, 5,. We de ne the Legendre symbol to be a p = 8 >< >: 1 if ais a quadratic residue modulo p 0 if pja 1 otherwise: This de nes a Dirichlet character of modulus p. Using 1 Theore insteadm, but otherwise following Vinogradov's argument, I prove: THEOREM 2. The This algorithm ends quite quickly, since it is known that the smallest quadratic non-residue is $<\sqrt{p}+1$, and if the extended Riemann hypothesis is true, then the smallest quadratic non-residue is $<\frac{3(\ln{p})^2}{2}$ N2 - Assuming the Generalized Riemann Hypothesis, it is known that the smallest quadratic non-residue modulo a prime p is less than or equal to (log p)2. We say that ais a quadratic residue mod pif ais a square mod p(it is a quadratic non-residue otherwise). Given a proper subgroup H of G, two natural and interesting questionsinnumbertheoryare: It is known that under the Generalized Riemann Hypothesis, the smallest quadratic non-residue modulo a prime p is less than or equal to (log p) 2. Now it is true that an upper bound 2p1/2 on the Quadratic non-residues, character sums. Theorem 3. Alternatively, let Nov 27, 2013 · Assuming the Generalized Riemann Hypothesis, it is known that the smallest quadratic non-residue modulo a prime p p is less than or equal to ( log ⁡ p ) 2 (\log p)^2 . Let m be the smallest positive integer such that m is a quadratic non-residue modulo p. Then we take the index of both sides to get ind rx2 ind ra mod p 1 and so 2ind rx ind a random prime is on average approximately equal to the 2k-th smallest prime. At least two of p 1, p 3, p 5 are non-residues, and at least one of these two is prime to u. 15163 … + o (1)) r. Finally we consider the least quadratic non-residue problem in Pajtechi˘ı-˘Sapiro’s sequence {[nc]}∞ n=1, where c>1 is a constant and [t] denotes the integral part of t∈ R. (The power of 1= p e is gained by sieving. Of course, all square numbers are trivially quadratic residues for all moduli. , the question of how big the least quadratic non-residue is, has been studied extensively. Crossref View in Scopus Google Scholar. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. . But if ais coprime to pthen the polynomial x2 a 0 mod p; either has two solutions or no solutions. Assuming the Generalized Riemann Hypothesis for Dirichlet L-functions, Ankeny [1] Preprint submitted to Elsevier August 29, 2014 In particular we prove the upper bound l p ≤ ( 0. =): Suppose a is a quadratic residue mod p so there exists some x with x2 a mod p. Aug 10, 2012 · I would not worry about it. So staring with small prines is a good strategy if only needing "a" quadratic non-residue. If such an xdoes not exist, we say that a is a quadratic non-residue. De nition. Stack Exchange Network. F. Use it to find the smallest prime p whose least quadratic non-residue is greater than 50 and compute the value of this quadratic non-residue. A QR m(mod n) is a non-zero QR if m6 0 (mod n). The case k= 2, i. May 2, 2019 · The justification for focusing on non-residues is typically that the least residue is trivial since 1 is a square for all integer moduli. This can be done by a straightforward modification of the standard "Euclid" proof, so for fun we show that there are infinitely many primes among the integers of $\mathbb{Q}[\sqrt{d}]$. But that means n p is a quadratic residue. We display the MQ instance induced by the evaluation of the sequential Legendre PRF, $\{5\}_K=(1,1,-1,-1,1)$ . everyoneloves__mid-leaderboard:empty,. 3. , gcd(a, p) = 1; a is called a quadratic residue if x2 = a mod p has a solution otherwise it is called a quadratic non-residue. Hint: Assume to the contrary and use Legendre Symbols the smallest quadratic non-residue (non-square) and the smallest primitive root (generator of F p) modulo p. If there is no quadratic non-residue in [1,N] then S χ(N) = N, so |S χ(N)| attains the trivial upper bound. Nov 12, 2009 · Show that the smallest quadratic non residue of an odd prime p is itself prime. Vinogradov to obtain an estimate from above of the smallest quadratic non-residue modulo $ p $. quadratic non-residue. Jul 31, 2021 · So, we can choose the smallest value from these two numbers i. That is, F = Z p[x]= x2 a; where a is any quadratic non-residue modulo p. Amer. More generally, one is interested in the distribution of subsets of Z=pZ de ned by algebraic conditions from F p for large values of the prime p. Here are all possible equations modulo p = 3,5 and 7, and whether each a is a quadratic residue modulo p. The proof of the Burgess theorem is based on an ampliﬂcation Mar 29, 2024 · These sums were used at the same time by I. Finally, we improve the best known theoretical bounds for the least quadratic non-residue, and more generally, the least k-th power non-residue. Hence, we have the trivial bound F(a,p) ≤ p2+3η0+o(1) for any a, which Write a Maple function that, given a positive integer n, finds the smallest prime p such that a is a quadratic residue modulo p for all 1 ≤ a ≤ n. It is elementary to get a bound of the form O(p p), which we will do, but much stronger results seem to be true. It is known that under the Generalized Riemann Hypothesis, the smallest quadratic non-residue modulo a prime $p$ is less than or equal to $(\\log p)^{2 k-th power non-residue mod p. Corollary 2 1 2 QR(q) , q 1 (mod 4) 1 2 QNR(q) , q 3 (mod 4): Observe that x2 = ( x)2, which, in the case of a prime eld Fp, implies that QR(p) = ˆ y2: 0 < y p 1 2 ˙: Let p be an odd prime. Lemma 39. 519-528. Oct 7, 2019 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have deduce explicit bounds for the class number of imaginary quadratic ﬁelds. Aug 19, 2019 · Stack Exchange Network. Legendre Symbol. Concerning the size of n´, P¶olya-Vinogradov Mar 29, 2022 · Stack Exchange Network. Then d = O(pa) as p -> oo, for any fixed a > ^e~1/2. Math. , 32 (1965), pp. Introduction Letq>1beanaturalnumberandletG=(Z/qZ)∗ denotethegroupofreduced residues (mod q). a equation solutions QR? 1 x2 ≡1 (mod 3) x ≡1,2 Nov 27, 2013 · We use the Burgess bound and combinatorial sieve to obtain an upper bound on the number of primes $p$ in a dyadic interval $[Q,2Q]$ for which a given interval $[u+1 p is the least quadratic non-residue, that means a;bare quadratic residues, but then m 2 1 amod pand m 2 bmod p, so (m 1m 2)2 ab= n pmod p. So you are looking for results where p = 1 or 7 (mod 8). 2) np ≤ pη for any η >η0 and a suﬃciently large p. Let q ‚ 2 be an integer and ´ a non principal Dirichlet character modulo q. This is joint work with Debmalya Basak and Alexandru Zaharescu. Noticing that an−3 p is a quadratic residue modulo p, we now obtain F(a,p) ≤ n3 p(p− 1)2. Then 2 is a quadratic residue modulo p if and only if p 1 (mod 8): PROOF Let F be the eld with p2 elements. Vinogradov conjectured that the least quadratic non-residue is $\ll p^{\epsilon}$ for any $\epsilon >0$ but this remains unknown. Prove that m is a prime. It turns out that the problem of least non-quadratic residues is controlled by estimates on L(1/2 + it, χ), where this L is the Dirichlet function described in the last paragraph. It seems that this conjecture received less attention than the one about the smallest quadratic non-residue. Note that we always assume" > 0 and p > c("). Let us consider the quadratic Dirichlet character $\chi(m):=\left(\frac{-n}{m}\right)$ whose conductor divides $4n$. p is the least quadratic non-residue, that means a;bare quadratic residues, but then m 2 1 amod pand m 2 bmod p, so (m 1m 2)2 ab= n pmod p. Prove q is a prime. Example 1. deduce explicit bounds for the class number of imaginary quadratic ﬁelds. Feb 17, 2017 · How would I find the least positive residue of say $6! \\bmod 7$ or $12! \\bmod 13$ I just learned modular arithmetic and my book doesn't explain what least positive residues are so I'm a bit lost. In practice, it suffices to restrict the range to 0<x<=|_p/2_|, where |_x_| is the floor function, because of the symmetry (p-x)^2=x^2 (mod p). In the context of factoring large semiprimes n=p*q, the number of quadratic non-residues (kronecker -1) and quadratic residues (kronecker +1) is identical. Let denote d the least positive quadratic non-residue (modp). Apr 1, 2015 · The case k = 2, i. , q r -th power non Jan 25, 2024 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Apr 29, 2013 · Stack Exchange Network. de; Date: Fri, 09 Jun 2023 11:26: The creation of unbreakable (in principle) codes often hinges on estimates such as these. This paper studies explicit and theoretical bounds for several interesting quantities in number theory, conditionally on the Generalized Riemann Hypothesis. Mar 31, 2025 · If there is no integer 0<x<p such that x^2=q (mod p), i. Jan 18, 2020 · Otherwise, a is said to be a quadratic non-residue modulo p. So what I know is that, since q is the smallest positive integer which is a quadratic non residue (mod p) then Legendre symbol (q|p) = -1 = q^((p-1)/2) (mod p). 1 0 and 1 are always quadratic residues mod n. 1. The smallest quadratic non-residue modulo p is at most p 1 4 p e +" for " > 0 and p > c("). Proof: (=: If ind ra is even then observe that r1 2 ind ra 2 a mod p and so a is a quadratic residue mod p. 2 you say is all you need, but p = 3 gives too low a bound. | Find, read and cite all 1 < c < b, and (b,p) = (c,p) = 1. Clearly this result is the best possible (in terms of β) until at least the Burgess bound [4, 5] on the smallest quadratic nonresidue is improved. We say that an integer mis a quadratic non-residue (QNR) mod nif it is not a quadratic residue. Aug 19, 2009 · (The first quadratic residue is, of course, ; the more interesting problem is the first quadratic non-residue. everyoneloves__top-leaderboard:empty,. If the congruence (35) does have a solution, then q is said to be a quadratic residue (mod p). The now-standard terminology for nonresidues can cause confusion. Let p be an odd prime and let q be the smallest positive integer which is a quadratic non residue (mod p). The usual tool for studying these examples is the character sum S ˜(x)= X 1 n x ˜(n) Sep 15, 2023 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Jul 24, 2023 · Abstract. For example, at this writing (January 2021) the Wikipedia page for a related concept used both ‘quadratic nonresidue’ and ‘non-quadratic residue’. V. Introduction. Harman [19, Theorem 3] also gives similar, albeit weaker, results for non cube-free It is known that under the Generalized Riemann Hypothesis, the smallest quadratic non-residue modulo a prime $p$ is less than or equal to $(\log p)^{2}$. For example, P´olya-Vinogradov lets us take N˛ p1/2 logp, giving pθ+o(1) for θ= 1/2. A famous result of Gauss 1 Nov 16, 2023 · For p = 3 or 5 (mod 8) the lowest non quadratic residue is 2. e, 8. Now it is true that an upper bound 2p1/2 on the QUADRATIC RESIDUE ANSD NON-RESIDUES 10. p 8n + 7. Hence any nontrivial bound on jS ˜(N)jproves the existence of a quadratic non-residue less than N. SHPARLINSKI Another Conjecture of Vinogradov is the bound d(p) = po(1) , where d(p) is the longest sequence of consecutive quadratic residues modulo p. The third applet takes a prime p as input, and provides the following output: a list of the quadratic residues modulo p; the smallest primitive root r modulo p; and, a table of the values of r 1, r 2, r 3, . and 7, at least one of t and t2 is a non-residue and either t/4 or t2/4 is an odd non-residue less than p/2. In ranges slightly larger, of size (log p) A with A > 2, we consider chains of r consecutive quadratic non-residues. M. Each consecutive Legendre symbol pairs define an equation. Jul 11, 2022 · Stack Exchange Network. In case ´ coincides with the Legendre symbol, n´ is a least quadratic non-residue. 高斯称它为算术中的宝石，他一人先后给出多个证明。 Aug 6, 2021 · Stack Exchange Network. Garaev [13] proved that for 1 <c<12 Nov 15, 2013 · Clearly the Burgess bound [3], [4] n p ≤ p 1 / 4 e 1 / 2 + o (1) on the smallest positive quadratic non-residue n p, which is always a prime number, immediately implies that the smallest Hamming weight ℓ p among prime quadratic non-residues modulo p satisfies ℓ p ≤ (1 4 e 1 / 2 + o (1)) r = (0. e. Oct 24, 2016 · You could also simply enumerate the primitive roots and quadratic (non)residues modulo $13$, for example, to gain some insight. 1 2 S. t. ) Probabilistic heuristics (presuming that each non-square integer has a 50-50 chance of being a quadratic residue) suggests that should have size , and indeed Vinogradov conjectured that for any . Thus precisely k residues classes are quadratic residues and so all of the roots of the polynomial xk 1 are quadratic Jul 7, 2018 · What is the smallest prime $p$ such that every prime $q < 40$ is a quadratic non-residue $\\pmod p$? Given that the probability that $q$ is a non-residue mod Jan 1, 2020 · F urthermore, since all elements of A are quadratic non-residues, the contribution to ( 3. It is conjectured that the smallest quadratic non-residue is bounded by O(log(p)2). This follows from index calculus. Assuming the Generalized Riemann Hypothesis for Dirichlet L-functions, Ankeny [1] showed that g ( p , 2 ) ≪ ( log ⁡ p ) 2 and Bach [2] made this explicit by proving (under GRH) that g ( p , 2 ) ≤ 2 ( log ⁡ p ) 2 . Otherwise it is a quadratic non-residue (QNR, or just NR). Mar 1, 2023 · The smallest quadratic non-residue modulo p is 2. The study of distribution of quadratic residues and quadratic non-residues modulo p has been considered with great interest in the literature (see for instance [1, 3,4,5,6,7, 10, 12, 13, 15,16,17,18,19,20,21,22,23,24,25]). k-th power non-residue mod p. We will prove this assuming the generalized 数论基本概念之一。它是初等数论中非常重要的结果，不仅可用来判断二次同余式是否有解，还有很多用途。C. between the multiplicative characters $ \chi = \chi ( m, p) $ and the additive Jan 29, 2019 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Gaussian phenomena for small quadratic residues and non-residues HTML articles powered by AMS MathViewer by Debmalya Basak, Kunjakanan Nath and Alexandru Zaharescu; Trans. Denote by nχ p0,c the least positive integer nsuch that [nc] is a quadratic non-residue (modp). Given any N, do you think you can ﬁnd a prime p with the smallest quadratic NON-residue > N? If d is the smallest quadratic non-residue modulo p then d = O(pa log2 p) where a = In 1927 Vinogradov [14], [15] proved THEOREM B. If k\p—l and if H 5 is the class of kth power residues or a class of k th power non-residues, modulo p, then the number of elements of H s that are t^x is x/k + Δ, where x p/h A transformation on the sum implies Dec 30, 2022 · PDF | We prove that the average of the $k$-th smallest prime quadratic non-residue modulo a prime approximates the $2k$-th smallest prime. Feb 19, 2024 · Finding patterns in quadratic residues and non-residues has been a subject in the number theory literature for a long time. Let p be an odd prime. It is still unknown whether the Corollary 1 The product of two quadratic residues or two non-residues is a quadratic residue, whereas the product of a residue and a non-residue gives a non-residue. . The Riemann hypothesis will give a very good bound on quadratic non-residues. Observe that ( 23) ( 1) 2 12 3 1 (mod 8); If there is no quadratic non-residue in [1;N] then S ˜(N) = N, so jS ˜(N)jattains the trivial upper bound. ) Corollary 1. Nov 7, 2023 · Stack Exchange Network. Inparticular, theonlyknownresult aboutd(p)isthebound d(p)≤ p1/4+o(1), which is due to Burgess [1] as well. The least quadratic non-residue modulo a fixed prime $q$ is also a prime, so just check the primes $2,3,5,7,11,13,17, \ldots$ in order by Fermat. large cube-free q, every reduced residue class modulo qcontains a qβ+ε-smooth positive integer s ≤ q9/4+ε. non-residues. Modulo 2, every integer is a quadratic residue. , r p–1, each reduced modulo p. Jul 7, 2024 · $\begingroup$ According to Eric Bach, Explicit bounds for primality testing and related problems, assuming ERH there is always a quadratic non-residue up to $2(\log n)^2$, which might be helpful $\endgroup$ – Nov 9, 2022 · Download Citation | Gaussian phenomena for small quadratic residues and non-residues | Assuming the Generalized Riemann Hypothesis, it is known that the smallest quadratic non-residue modulo a Apr 19, 2021 · Given a number a, s. Jan 26, 2015 · quadratic ﬁelds. 376 (2023), 3695-3724 N consecutive quadratic non-residues which furthermore are not primitive roots. nliam ybpam rol xtxm dzbyb qrdwds ewa ffoavq ewj zhnbz xca wuufd mwvippy tdqfxm urhdwxg