Sums of two squares in arithmetic progressionsSums of two squares in (certain) integral domainsIs there another simple formula for the sum-of-squares function?Exact formula for the number of integers in an interval which are the sum of two squares.Arithmetic Progressions of SquaresSums of two squares: What is known about the distribution of r(n)?Primes in arithmetic progressions in number fieldsJacobi's theorem on sums of two squares (reference request)Average class number formula for imaginary quadratic fields with prime discriminantSums of four coprime squaresSums of two integer squares in arithmetic progressions

Sums of two squares in arithmetic progressions


Sums of two squares in (certain) integral domainsIs there another simple formula for the sum-of-squares function?Exact formula for the number of integers in an interval which are the sum of two squares.Arithmetic Progressions of SquaresSums of two squares: What is known about the distribution of r(n)?Primes in arithmetic progressions in number fieldsJacobi's theorem on sums of two squares (reference request)Average class number formula for imaginary quadratic fields with prime discriminantSums of four coprime squaresSums of two integer squares in arithmetic progressions













9












$begingroup$


Let $r(n)$ denote the number of representations of $n$ as the sum of two squares. Are there any known results on $$sum _nleq xatop nequiv a(q)r(n)$$ and in particular is there an asymptotic formula?










share|cite|improve this question











$endgroup$











  • $begingroup$
    If we fix $a, q$ and increase $x$, isn't this essentially the Gauss circle problem with some shift in the lattice?
    $endgroup$
    – Dongryul Kim
    2 days ago










  • $begingroup$
    It would be nice to make clear whether you require uniformity in $a$ and $q$. If yes, then the answer of Ofir addresses this. If not, then the result is a simple application of Dirichlet series techniques.
    $endgroup$
    – Daniel Loughran
    2 days ago















9












$begingroup$


Let $r(n)$ denote the number of representations of $n$ as the sum of two squares. Are there any known results on $$sum _nleq xatop nequiv a(q)r(n)$$ and in particular is there an asymptotic formula?










share|cite|improve this question











$endgroup$











  • $begingroup$
    If we fix $a, q$ and increase $x$, isn't this essentially the Gauss circle problem with some shift in the lattice?
    $endgroup$
    – Dongryul Kim
    2 days ago










  • $begingroup$
    It would be nice to make clear whether you require uniformity in $a$ and $q$. If yes, then the answer of Ofir addresses this. If not, then the result is a simple application of Dirichlet series techniques.
    $endgroup$
    – Daniel Loughran
    2 days ago













9












9








9


1



$begingroup$


Let $r(n)$ denote the number of representations of $n$ as the sum of two squares. Are there any known results on $$sum _nleq xatop nequiv a(q)r(n)$$ and in particular is there an asymptotic formula?










share|cite|improve this question











$endgroup$




Let $r(n)$ denote the number of representations of $n$ as the sum of two squares. Are there any known results on $$sum _nleq xatop nequiv a(q)r(n)$$ and in particular is there an asymptotic formula?







nt.number-theory reference-request analytic-number-theory arithmetic-progression sums-of-squares






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









GH from MO

59.2k5148227




59.2k5148227










asked 2 days ago









cawscaws

734




734











  • $begingroup$
    If we fix $a, q$ and increase $x$, isn't this essentially the Gauss circle problem with some shift in the lattice?
    $endgroup$
    – Dongryul Kim
    2 days ago










  • $begingroup$
    It would be nice to make clear whether you require uniformity in $a$ and $q$. If yes, then the answer of Ofir addresses this. If not, then the result is a simple application of Dirichlet series techniques.
    $endgroup$
    – Daniel Loughran
    2 days ago
















  • $begingroup$
    If we fix $a, q$ and increase $x$, isn't this essentially the Gauss circle problem with some shift in the lattice?
    $endgroup$
    – Dongryul Kim
    2 days ago










  • $begingroup$
    It would be nice to make clear whether you require uniformity in $a$ and $q$. If yes, then the answer of Ofir addresses this. If not, then the result is a simple application of Dirichlet series techniques.
    $endgroup$
    – Daniel Loughran
    2 days ago















$begingroup$
If we fix $a, q$ and increase $x$, isn't this essentially the Gauss circle problem with some shift in the lattice?
$endgroup$
– Dongryul Kim
2 days ago




$begingroup$
If we fix $a, q$ and increase $x$, isn't this essentially the Gauss circle problem with some shift in the lattice?
$endgroup$
– Dongryul Kim
2 days ago












$begingroup$
It would be nice to make clear whether you require uniformity in $a$ and $q$. If yes, then the answer of Ofir addresses this. If not, then the result is a simple application of Dirichlet series techniques.
$endgroup$
– Daniel Loughran
2 days ago




$begingroup$
It would be nice to make clear whether you require uniformity in $a$ and $q$. If yes, then the answer of Ofir addresses this. If not, then the result is a simple application of Dirichlet series techniques.
$endgroup$
– Daniel Loughran
2 days ago










1 Answer
1






active

oldest

votes


















16












$begingroup$

The first result in this direction seems to be due to R. A. Smith, ``The Circle Problem in an Arithmetic Progression,'' Can. Math. Bull. 11 (2), 175–184 (1968). He showed that if we write
$$sum _nleq xatop nequiv a(q)r(n) =pi x cdot fraceta_a(q)q^2+ R_q,a(x)$$
where $eta_a(q) := (x_1,x_2) in (mathbbZ/qmathbbZ)^2 : x_1^2 +x_2^2 equiv a bmod q$, then
$$R_q,a(x) = Oleft( x^frac23 + xi q^-frac12(1+3xi)gcd(a,q)^1/2tau(q) right)$$
for any $xi in (0,1/3)$. This is non-trivial for $q le x^frac23-varepsilon$. In particular, as $x$ tends to infinity, your expression is asymptotic to $pi x$ times the probability that $x_1^2+x_2^2 equiv a bmod q$ (for uniformly drawn $x_1,x_2$ mod $q$). If you consider $a$ and $q$ as fixed, this answers your question.



The state-of-the-art result is due to D. I. Tolev, ``On the remainder term in the circle problem in an arithmetic progression,'' Tr. Mat. Inst. Steklova 276 (2012), Teoriya Chisel, Algebra i Analiz, 266--279; translation in Proc. Steklov Inst. Math. 276 (2012), no. 1, 261–274. He showed that
$$R_q,a(x) = Oleft( (q^frac12+x^frac13) gcd(a,q)^1/2tau^4(q)log^4 x right).$$
Interestingly, for $a=1$, there is a result which is superior in certain ranges of $x$ and $q$, see P. D. Varbanets, “Lattice Points in a Circle Whose Distances from the Center Are in an Arithmetic Progression,” Mat. Zametki 8 (6), 787–798 (1970) [Math. Notes 8, 917–923 (1970)].



All three results are explained in Tolev's paper.




In the 2002 PhD thesis of Michael J. Dancs under R. Vaughn, titled "On a Variance arising in the Gauss Circle Problem", he proves (independently of the above) that
$$f(q,a):=lim_x to infty fracsum _nleq xatop nequiv a(q)r(n) pi x$$
exists and can be written as
$$f(q,a)=q^-3 sum_k=1^q expleft( 2pi i frac-akq right) S(q,k)^2,$$
where $S(q,a)$ is a quadratic Gauss sum mod $q$. This gives a different expression for $eta_a(q)/q^2$, which was better suited for the applications given by Dancs. This is done in Theorem 2.1. He then proceeds to prove, in Theorem 2.2, the estimate
$$R_q,a(x) = Oleft( (sqrtx+q) log qright).$$
For small $q$ this is better than Smith's work. Still, it is superseded by Tolev's estimates.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks for the very informative answer!
    $endgroup$
    – caws
    2 days ago











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1 Answer
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1 Answer
1






active

oldest

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active

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active

oldest

votes









16












$begingroup$

The first result in this direction seems to be due to R. A. Smith, ``The Circle Problem in an Arithmetic Progression,'' Can. Math. Bull. 11 (2), 175–184 (1968). He showed that if we write
$$sum _nleq xatop nequiv a(q)r(n) =pi x cdot fraceta_a(q)q^2+ R_q,a(x)$$
where $eta_a(q) := (x_1,x_2) in (mathbbZ/qmathbbZ)^2 : x_1^2 +x_2^2 equiv a bmod q$, then
$$R_q,a(x) = Oleft( x^frac23 + xi q^-frac12(1+3xi)gcd(a,q)^1/2tau(q) right)$$
for any $xi in (0,1/3)$. This is non-trivial for $q le x^frac23-varepsilon$. In particular, as $x$ tends to infinity, your expression is asymptotic to $pi x$ times the probability that $x_1^2+x_2^2 equiv a bmod q$ (for uniformly drawn $x_1,x_2$ mod $q$). If you consider $a$ and $q$ as fixed, this answers your question.



The state-of-the-art result is due to D. I. Tolev, ``On the remainder term in the circle problem in an arithmetic progression,'' Tr. Mat. Inst. Steklova 276 (2012), Teoriya Chisel, Algebra i Analiz, 266--279; translation in Proc. Steklov Inst. Math. 276 (2012), no. 1, 261–274. He showed that
$$R_q,a(x) = Oleft( (q^frac12+x^frac13) gcd(a,q)^1/2tau^4(q)log^4 x right).$$
Interestingly, for $a=1$, there is a result which is superior in certain ranges of $x$ and $q$, see P. D. Varbanets, “Lattice Points in a Circle Whose Distances from the Center Are in an Arithmetic Progression,” Mat. Zametki 8 (6), 787–798 (1970) [Math. Notes 8, 917–923 (1970)].



All three results are explained in Tolev's paper.




In the 2002 PhD thesis of Michael J. Dancs under R. Vaughn, titled "On a Variance arising in the Gauss Circle Problem", he proves (independently of the above) that
$$f(q,a):=lim_x to infty fracsum _nleq xatop nequiv a(q)r(n) pi x$$
exists and can be written as
$$f(q,a)=q^-3 sum_k=1^q expleft( 2pi i frac-akq right) S(q,k)^2,$$
where $S(q,a)$ is a quadratic Gauss sum mod $q$. This gives a different expression for $eta_a(q)/q^2$, which was better suited for the applications given by Dancs. This is done in Theorem 2.1. He then proceeds to prove, in Theorem 2.2, the estimate
$$R_q,a(x) = Oleft( (sqrtx+q) log qright).$$
For small $q$ this is better than Smith's work. Still, it is superseded by Tolev's estimates.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks for the very informative answer!
    $endgroup$
    – caws
    2 days ago















16












$begingroup$

The first result in this direction seems to be due to R. A. Smith, ``The Circle Problem in an Arithmetic Progression,'' Can. Math. Bull. 11 (2), 175–184 (1968). He showed that if we write
$$sum _nleq xatop nequiv a(q)r(n) =pi x cdot fraceta_a(q)q^2+ R_q,a(x)$$
where $eta_a(q) := (x_1,x_2) in (mathbbZ/qmathbbZ)^2 : x_1^2 +x_2^2 equiv a bmod q$, then
$$R_q,a(x) = Oleft( x^frac23 + xi q^-frac12(1+3xi)gcd(a,q)^1/2tau(q) right)$$
for any $xi in (0,1/3)$. This is non-trivial for $q le x^frac23-varepsilon$. In particular, as $x$ tends to infinity, your expression is asymptotic to $pi x$ times the probability that $x_1^2+x_2^2 equiv a bmod q$ (for uniformly drawn $x_1,x_2$ mod $q$). If you consider $a$ and $q$ as fixed, this answers your question.



The state-of-the-art result is due to D. I. Tolev, ``On the remainder term in the circle problem in an arithmetic progression,'' Tr. Mat. Inst. Steklova 276 (2012), Teoriya Chisel, Algebra i Analiz, 266--279; translation in Proc. Steklov Inst. Math. 276 (2012), no. 1, 261–274. He showed that
$$R_q,a(x) = Oleft( (q^frac12+x^frac13) gcd(a,q)^1/2tau^4(q)log^4 x right).$$
Interestingly, for $a=1$, there is a result which is superior in certain ranges of $x$ and $q$, see P. D. Varbanets, “Lattice Points in a Circle Whose Distances from the Center Are in an Arithmetic Progression,” Mat. Zametki 8 (6), 787–798 (1970) [Math. Notes 8, 917–923 (1970)].



All three results are explained in Tolev's paper.




In the 2002 PhD thesis of Michael J. Dancs under R. Vaughn, titled "On a Variance arising in the Gauss Circle Problem", he proves (independently of the above) that
$$f(q,a):=lim_x to infty fracsum _nleq xatop nequiv a(q)r(n) pi x$$
exists and can be written as
$$f(q,a)=q^-3 sum_k=1^q expleft( 2pi i frac-akq right) S(q,k)^2,$$
where $S(q,a)$ is a quadratic Gauss sum mod $q$. This gives a different expression for $eta_a(q)/q^2$, which was better suited for the applications given by Dancs. This is done in Theorem 2.1. He then proceeds to prove, in Theorem 2.2, the estimate
$$R_q,a(x) = Oleft( (sqrtx+q) log qright).$$
For small $q$ this is better than Smith's work. Still, it is superseded by Tolev's estimates.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks for the very informative answer!
    $endgroup$
    – caws
    2 days ago













16












16








16





$begingroup$

The first result in this direction seems to be due to R. A. Smith, ``The Circle Problem in an Arithmetic Progression,'' Can. Math. Bull. 11 (2), 175–184 (1968). He showed that if we write
$$sum _nleq xatop nequiv a(q)r(n) =pi x cdot fraceta_a(q)q^2+ R_q,a(x)$$
where $eta_a(q) := (x_1,x_2) in (mathbbZ/qmathbbZ)^2 : x_1^2 +x_2^2 equiv a bmod q$, then
$$R_q,a(x) = Oleft( x^frac23 + xi q^-frac12(1+3xi)gcd(a,q)^1/2tau(q) right)$$
for any $xi in (0,1/3)$. This is non-trivial for $q le x^frac23-varepsilon$. In particular, as $x$ tends to infinity, your expression is asymptotic to $pi x$ times the probability that $x_1^2+x_2^2 equiv a bmod q$ (for uniformly drawn $x_1,x_2$ mod $q$). If you consider $a$ and $q$ as fixed, this answers your question.



The state-of-the-art result is due to D. I. Tolev, ``On the remainder term in the circle problem in an arithmetic progression,'' Tr. Mat. Inst. Steklova 276 (2012), Teoriya Chisel, Algebra i Analiz, 266--279; translation in Proc. Steklov Inst. Math. 276 (2012), no. 1, 261–274. He showed that
$$R_q,a(x) = Oleft( (q^frac12+x^frac13) gcd(a,q)^1/2tau^4(q)log^4 x right).$$
Interestingly, for $a=1$, there is a result which is superior in certain ranges of $x$ and $q$, see P. D. Varbanets, “Lattice Points in a Circle Whose Distances from the Center Are in an Arithmetic Progression,” Mat. Zametki 8 (6), 787–798 (1970) [Math. Notes 8, 917–923 (1970)].



All three results are explained in Tolev's paper.




In the 2002 PhD thesis of Michael J. Dancs under R. Vaughn, titled "On a Variance arising in the Gauss Circle Problem", he proves (independently of the above) that
$$f(q,a):=lim_x to infty fracsum _nleq xatop nequiv a(q)r(n) pi x$$
exists and can be written as
$$f(q,a)=q^-3 sum_k=1^q expleft( 2pi i frac-akq right) S(q,k)^2,$$
where $S(q,a)$ is a quadratic Gauss sum mod $q$. This gives a different expression for $eta_a(q)/q^2$, which was better suited for the applications given by Dancs. This is done in Theorem 2.1. He then proceeds to prove, in Theorem 2.2, the estimate
$$R_q,a(x) = Oleft( (sqrtx+q) log qright).$$
For small $q$ this is better than Smith's work. Still, it is superseded by Tolev's estimates.






share|cite|improve this answer











$endgroup$



The first result in this direction seems to be due to R. A. Smith, ``The Circle Problem in an Arithmetic Progression,'' Can. Math. Bull. 11 (2), 175–184 (1968). He showed that if we write
$$sum _nleq xatop nequiv a(q)r(n) =pi x cdot fraceta_a(q)q^2+ R_q,a(x)$$
where $eta_a(q) := (x_1,x_2) in (mathbbZ/qmathbbZ)^2 : x_1^2 +x_2^2 equiv a bmod q$, then
$$R_q,a(x) = Oleft( x^frac23 + xi q^-frac12(1+3xi)gcd(a,q)^1/2tau(q) right)$$
for any $xi in (0,1/3)$. This is non-trivial for $q le x^frac23-varepsilon$. In particular, as $x$ tends to infinity, your expression is asymptotic to $pi x$ times the probability that $x_1^2+x_2^2 equiv a bmod q$ (for uniformly drawn $x_1,x_2$ mod $q$). If you consider $a$ and $q$ as fixed, this answers your question.



The state-of-the-art result is due to D. I. Tolev, ``On the remainder term in the circle problem in an arithmetic progression,'' Tr. Mat. Inst. Steklova 276 (2012), Teoriya Chisel, Algebra i Analiz, 266--279; translation in Proc. Steklov Inst. Math. 276 (2012), no. 1, 261–274. He showed that
$$R_q,a(x) = Oleft( (q^frac12+x^frac13) gcd(a,q)^1/2tau^4(q)log^4 x right).$$
Interestingly, for $a=1$, there is a result which is superior in certain ranges of $x$ and $q$, see P. D. Varbanets, “Lattice Points in a Circle Whose Distances from the Center Are in an Arithmetic Progression,” Mat. Zametki 8 (6), 787–798 (1970) [Math. Notes 8, 917–923 (1970)].



All three results are explained in Tolev's paper.




In the 2002 PhD thesis of Michael J. Dancs under R. Vaughn, titled "On a Variance arising in the Gauss Circle Problem", he proves (independently of the above) that
$$f(q,a):=lim_x to infty fracsum _nleq xatop nequiv a(q)r(n) pi x$$
exists and can be written as
$$f(q,a)=q^-3 sum_k=1^q expleft( 2pi i frac-akq right) S(q,k)^2,$$
where $S(q,a)$ is a quadratic Gauss sum mod $q$. This gives a different expression for $eta_a(q)/q^2$, which was better suited for the applications given by Dancs. This is done in Theorem 2.1. He then proceeds to prove, in Theorem 2.2, the estimate
$$R_q,a(x) = Oleft( (sqrtx+q) log qright).$$
For small $q$ this is better than Smith's work. Still, it is superseded by Tolev's estimates.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Ofir GorodetskyOfir Gorodetsky

5,90312639




5,90312639











  • $begingroup$
    Thanks for the very informative answer!
    $endgroup$
    – caws
    2 days ago
















  • $begingroup$
    Thanks for the very informative answer!
    $endgroup$
    – caws
    2 days ago















$begingroup$
Thanks for the very informative answer!
$endgroup$
– caws
2 days ago




$begingroup$
Thanks for the very informative answer!
$endgroup$
– caws
2 days ago

















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Cannot Extend partition with GParted The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Community Moderator Election ResultsCan't increase partition size with GParted?GParted doesn't recognize the unallocated space after my current partitionWhat is the best way to add unallocated space located before to Ubuntu 12.04 partition with GParted live?I can't figure out how to extend my Arch home partition into free spaceGparted Linux Mint 18.1 issueTrying to extend but swap partition is showing as Unknown in Gparted, shows proper from fdiskRearrange partitions in gparted to extend a partitionUnable to extend partition even though unallocated space is next to it using GPartedAllocate free space to root partitiongparted: how to merge unallocated space with a partition

대한민국 목차 국명 지리 역사 정치 국방 경제 사회 문화 국제 순위 관련 항목 각주 외부 링크 둘러보기 메뉴북위 37° 34′ 08″ 동경 126° 58′ 36″ / 북위 37.568889° 동경 126.976667°  / 37.568889; 126.976667ehThe Korean Repository문단을 편집문단을 편집추가해Clarkson PLC 사Report for Selected Countries and Subjects-Korea“Human Development Index and its components: P.198”“http://www.law.go.kr/%EB%B2%95%EB%A0%B9/%EB%8C%80%ED%95%9C%EB%AF%BC%EA%B5%AD%EA%B5%AD%EA%B8%B0%EB%B2%95”"한국은 국제법상 한반도 유일 합법정부 아니다" - 오마이뉴스 모바일Report for Selected Countries and Subjects: South Korea격동의 역사와 함께한 조선일보 90년 : 조선일보 인수해 혁신시킨 신석우, 임시정부 때는 '대한민국' 국호(國號) 정해《우리가 몰랐던 우리 역사: 나라 이름의 비밀을 찾아가는 역사 여행》“남북 공식호칭 ‘남한’‘북한’으로 쓴다”“Corea 대 Korea, 누가 이긴 거야?”국내기후자료 - 한국[김대중 前 대통령 서거] 과감한 구조개혁 'DJ노믹스'로 최단기간 환란극복 :: 네이버 뉴스“이라크 "韓-쿠르드 유전개발 MOU 승인 안해"(종합)”“해외 우리국민 추방사례 43%가 일본”차기전차 K2'흑표'의 세계 최고 전력 분석, 쿠키뉴스 엄기영, 2007-03-02두산인프라, 헬기잡는 장갑차 'K21'...내년부터 공급, 고뉴스 이대준, 2008-10-30과거 내용 찾기mk 뉴스 - 구매력 기준으로 보면 한국 1인당 소득 3만弗과거 내용 찾기"The N-11: More Than an Acronym"Archived조선일보 최우석, 2008-11-01Global 500 2008: Countries - South Korea“몇년째 '시한폭탄'... 가계부채, 올해는 터질까”가구당 부채 5000만원 처음 넘어서“‘빚’으로 내몰리는 사회.. 위기의 가계대출”“[경제365] 공공부문 부채 급증…800조 육박”“"소득 양극화 다소 완화...불평등은 여전"”“공정사회·공생발전 한참 멀었네”iSuppli,08年2QのDRAMシェア・ランキングを発表(08/8/11)South Korea dominates shipbuilding industry | Stock Market News & Stocks to Watch from StraightStocks한국 자동차 생산, 3년 연속 세계 5위자동차수출 '현대-삼성 웃고 기아-대우-쌍용은 울고' 과거 내용 찾기동반성장위 창립 1주년 맞아Archived"중기적합 3개업종 합의 무시한 채 선정"李대통령, 사업 무분별 확장 소상공인 생계 위협 질타삼성-LG, 서민업종인 빵·분식사업 잇따라 철수상생은 뒷전…SSM ‘몸집 불리기’ 혈안Archived“경부고속도에 '아시안하이웨이' 표지판”'철의 실크로드' 앞서 '말(言)의 실크로드'부터, 프레시안 정창현, 2008-10-01“'서울 지하철은 안전한가?'”“서울시 “올해 안에 모든 지하철역 스크린도어 설치””“부산지하철 1,2호선 승강장 안전펜스 설치 완료”“전교조, 정부 노조 통계서 처음 빠져”“[Weekly BIZ] 도요타 '제로 이사회'가 리콜 사태 불러들였다”“S Korea slams high tuition costs”““정치가 여론 양극화 부채질… 합리주의 절실””“〈"`촛불집회'는 민주주의의 질적 변화 상징"〉”““촛불집회가 민주주의 왜곡 초래””“국민 65%, "한국 노사관계 대립적"”“한국 국가경쟁력 27위‥노사관계 '꼴찌'”“제대로 형성되지 않은 대한민국 이념지형”“[신년기획-갈등의 시대] 갈등지수 OECD 4위…사회적 손실 GDP 27% 무려 300조”“2012 총선-대선의 키워드는 '국민과 소통'”“한국 삶의 질 27위, 2000년과 2008년 연속 하위권 머물러”“[해피 코리아] 행복점수 68점…해외 평가선 '낙제점'”“한국 어린이·청소년 행복지수 3년 연속 OECD ‘꼴찌’”“한국 이혼율 OECD중 8위”“[통계청] 한국 이혼율 OECD 4위”“오피니언 [이렇게 생각한다] `부부의 날` 에 돌아본 이혼율 1위 한국”“Suicide Rates by Country, Global Health Observatory Data Repository.”“1. 또 다른 차별”“오피니언 [편집자에게] '왕따'와 '패거리 정치' 심리는 닮은꼴”“[미래한국리포트] 무한경쟁에 빠진 대한민국”“대학생 98% "외모가 경쟁력이라는 말 동의"”“특급호텔 웨딩·200만원대 유모차… "남보다 더…" 호화病, 고질병 됐다”“[스트레스 공화국] ① 경쟁사회, 스트레스 쌓인다”““매일 30여명 자살 한국, 의사보다 무속인에…””“"자살 부르는 '우울증', 환자 중 85% 치료 안 받아"”“정신병원을 가다”“대한민국도 ‘묻지마 범죄’,안전지대 아니다”“유엔 "학생 '성적 지향'에 따른 차별 금지하라"”“유엔아동권리위원회 보고서 및 번역본 원문”“고졸 성공스토리 담은 '제빵왕 김탁구' 드라마 나온다”“‘빛 좋은 개살구’ 고졸 취업…실습 대신 착취”원본 문서“정신건강, 사회적 편견부터 고쳐드립니다”‘소통’과 ‘행복’에 목 마른 사회가 잠들어 있던 ‘심리학’ 깨웠다“[포토] 사유리-곽금주 교수의 유쾌한 심리상담”“"올해 한국인 평균 영화관람횟수 세계 1위"(종합)”“[게임연중기획] 게임은 문화다-여가활동 1순위 게임”“영화속 ‘영어 지상주의’ …“왠지 씁쓸한데””“2월 `신문 부수 인증기관` 지정..방송법 후속작업”“무료신문 성장동력 ‘차별성’과 ‘갈등해소’”대한민국 국회 법률지식정보시스템"Pew Research Center's Religion & Public Life Project: South Korea"“amp;vwcd=MT_ZTITLE&path=인구·가구%20>%20인구총조사%20>%20인구부문%20>%20 총조사인구(2005)%20>%20전수부문&oper_YN=Y&item=&keyword=종교별%20인구& amp;lang_mode=kor&list_id= 2005년 통계청 인구 총조사”원본 문서“한국인이 좋아하는 취미와 운동 (2004-2009)”“한국인이 좋아하는 취미와 운동 (2004-2014)”Archived“한국, `부분적 언론자유국' 강등〈프리덤하우스〉”“국경없는기자회 "한국, 인터넷감시 대상국"”“한국, 조선산업 1위 유지(S. Korea Stays Top Shipbuilding Nation) RZD-Partner Portal”원본 문서“한국, 4년 만에 ‘선박건조 1위’”“옛 마산시,인터넷속도 세계 1위”“"한국 초고속 인터넷망 세계1위"”“인터넷·휴대폰 요금, 외국보다 훨씬 비싸”“한국 관세행정 6년 연속 세계 '1위'”“한국 교통사고 사망자 수 OECD 회원국 중 2위”“결핵 후진국' 한국, 환자가 급증한 이유는”“수술은 신중해야… 자칫하면 생명 위협”대한민국분류대한민국의 지도대한민국 정부대표 다국어포털대한민국 전자정부대한민국 국회한국방송공사about korea and information korea브리태니커 백과사전(한국편)론리플래닛의 정보(한국편)CIA의 세계 정보(한국편)마리암 부디아 (Mariam Budia),『한국: 하늘이 내린 한 폭의 그림』, 서울: 트랜스라틴 19호 (2012년 3월)대한민국ehehehehehehehehehehehehehehWorldCat132441370n791268020000 0001 2308 81034078029-6026373548cb11863345f(데이터)00573706ge128495