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Which $n$ can let $S=1+\frac14+\frac19+\cdots+\frac1{n^2}$ be a square of a rational number? Obviously, $1$ and $3$ work, but how to prove they are the only ones?

I think this problem is really hard. I have asked many professional number theorists in a top-5 math department in US, and none of them can give any clue. I am thinking this could be an open question.

If you cannot solve this, try this: let $S=1+\frac12+\frac13+\cdots+\frac1{n}$. Which n can make this an integer. Prove your result. This is proposed by a famous mathematician Shing Tung Yau for a college math competition.

Hopefully someone can give a clue to No.1. I suggested if you cannot solve the first one but would like to see an answer, rate this problem up so that it may have a better chance to bring in more experts for a solution. Thanks!

It is also appreciated if you would like to post your thoughts on Yau's problem!

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  • $\begingroup$ Where did you find this question? Or is it your own creation? $\endgroup$
    – Potato
    Oct 30, 2014 at 1:08
  • $\begingroup$ originally in a math forum specifically for math olympiad in Asia, where people post their unsolved question for a discussion. And it turns out that no one can solve this one. Maybe an open question $\endgroup$
    – gter
    Oct 30, 2014 at 1:12
  • $\begingroup$ Is the $($normal$)$ harmonic number $H_n$ ever a rational square for $n>1$? If not, has this been proven? $\endgroup$
    – Lucian
    Oct 31, 2014 at 17:04
  • $\begingroup$ The second part is easy. See here for a hint. But I really wanted to draw your attention to Keith Conrad's remark under my answer. He gives a related open problem there! $\endgroup$ Nov 1, 2014 at 21:12

3 Answers 3

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I do not see anything intuitive about the first problem, so I put it on computer. I normalized by having the denominator at step $n$ be $(n!)^2.$ A human being would probably use $\operatorname{lcm}(1,2,3,\ldots,n)^2 $ which is much smaller, but the computer does not care.

There are some reasons to expect the conjecture to be true. The primes $2,3,7$ are special. For the other primes, $5,11,13,17,19,...$ the following pattern happens and can probably be proved: an odd power of the prime divides the numerator for $$ n = (p-1)/2, \; \; n = (p-1), $$ $$ p (p-1)/2 \leq n \leq p (p+1)/2 - 1, \; \; \; p (p-1) \leq n \leq p^2 - 1, $$ $$ p^2 (p-1)/2 \leq n \leq p^2 (p+1)/2 - 1, \; \; \; p^2 (p-1) \leq n \leq p^3 - 1, $$ $$ p^3 (p-1)/2 \leq n \leq p^3 (p+1)/2 - 1, \; \; \; p^3 (p-1) \leq n \leq p^4 - 1, $$ Evidently the exponents for $2$ and $3$ are always even. The pattern for odd exponents of $7$ is of the same density as the larger primes but the beginnings are a little off: $6,26; 42-28,182-188; 294-342,1274-1322,$ and so on. Multiply each interval starting value by $7$ and the interval length by $7.$

 October 31 

2  5 = .5
3  49 = 7^2
4  820 = 2^2.5 .41
5  21076 = 2^2.11 .479
6  773136 = 2^4.3^2.7.13 .59
7  38402064 = 2^4.3^2 .266681
8  2483133696 = 2^8.3^2.17 .63397
9  202759531776 = 2^8.3^4.19 .514639
10  20407635072000 = 2^10.3^4.5^3.11 .178939
11  2482492033152000 = 2^10.3^4.5^3.23.43 .242101
12  359072203696128000 = 2^14.3^6.5^3.13 .18500393
13  60912644957448192000 = 2^14.3^6.5^3 . mbox{BIG} 
14  11977654199703478272000 = 2^16.3^6.5^3.7^2.29.7417 .190297
15  2702572249389834608640000 = 2^16.3^8.5^4.7^2.31.37.97 .1844659
16  693568508096521859235840000 = 2^22.3^8.5^4.7^2.17.619 .78206663
17  200879061976592212371701760000 = 2^22.3^8.5^4.7^2.601.643 .616798327
18  65211329626921423978616586240000 = 2^24.3^12.5^4.7^2.19.37.8821 .38512247
19  23582280384386431339420597616640000 = 2^24.3^12.5^4.7^2 . mbox{BIG} 
20  9447709684208047354981782650880000000 = 2^28.3^12.5^7.7^2.41 . mbox{BIG} 
21  4172358982917138811232383590727680000000 = 2^28.3^14.5^7.7^6.37.43.2621 .84786899
22  2022032032103888142745742749697310720000000 = 2^30.3^14.5^7.7^6.11^2.23.295831 .52030193
23  1070918322619001419237384155013872353280000000 = 2^30.3^14.5^7.7^6.11^2.47.127 . mbox{BIG} 
24  617517280598012406503198094472283821178880000000 = 2^36.3^16.5^7.7^6.11^2.59.2237 . mbox{BIG} 
25  386333256592971085341438546047330355354009600000000 = 2^36.3^16.5^8.7^6.11^2.157.3119 . mbox{BIG} 
26  261401879093856785738899792754340924667566489600000000 = 2^38.3^16.5^8.7^7.11^2.13^2.53.70853.106357 .8408339
27  190724613862039229268164987801324162689677026918400000000 = 2^38.3^20.5^8.7^6.11^2.13^2.307.110023 . mbox{BIG} 
jagy@phobeusjunior

What I did next had everything to do with the patterns for odd exponents. I did a limited factoring of eache denominator, divisibility by just the primes up to $2n+1.$ As a result, there was generally some large factor $F$ remaining, probably composite but no matter. Then I just asked for $F \bmod 8,$ then the first four primes $q$ such that the Jacobi symbol $(F|q) = -1.$ These give a more satisfying feel than simply asking the computer whether $F$ is a square, they give a specific reason for it to fail to be a square. Oh, to save space, I stopped printing out the denominator itself and stopped printing out the (many) prime factors with even exponents. These became quite numerous, for $p = 5,11,13,17,...$ the prime exponents always stayed the same or increased for $n \geq 2p.$ I put this next list up to 74 so you could see how the exponent of 5 stayed odd for $50 \leq n \leq 74$ by increasing by 2.

2 = .5
3 = 
4 = .5 .41
5 = .11 . mbox{big non-square}      8: 7  J 3  J 7  J 11  J 13
6 = .7.13 .59
7 =  . mbox{big non-square}      8: 1  J 3  J 11  J 23  J 29
8 = .17 . mbox{big non-square}      8: 5  J 5  J 7  J 19  J 29
9 = .19 . mbox{big non-square}      8: 7  J 7  J 13  J 23  J 37
10 = .5^3.11 . mbox{big non-square}      8: 3  J 7  J 11  J 13  J 17
11 = .5^3.23 . mbox{big non-square}      8: 7  J 5  J 7  J 13  J 19
12 = .5^3.13 . mbox{big non-square}      8: 1  J 3  J 5  J 11  J 13
13 = .5^3 . mbox{big non-square}      8: 5  J 3  J 7  J 13  J 17
14 = .5^3.29 . mbox{big non-square}      8: 1  J 11  J 13  J 17  J 19
15 = .31 . mbox{big non-square}      8: 7  J 13  J 17  J 29  J 31
16 = .17 . mbox{big non-square}      8: 5  J 3  J 5  J 11  J 17
17 =  . mbox{big non-square}      8: 5  J 7  J 19  J 23  J 29
18 = .19.37 . mbox{big non-square}      8: 3  J 3  J 5  J 11  J 17
19 =  . mbox{big non-square}      8: 5  J 3  J 7  J 23  J 29
20 = .5^7.41 . mbox{big non-square}      8: 1  J 3  J 7  J 11  J 19
21 = .5^7.37.43 . mbox{big non-square}      8: 7  J 7  J 11  J 19  J 29
22 = .5^7.23 . mbox{big non-square}      8: 7  J 3  J 5  J 7  J 13
23 = .5^7.47 . mbox{big non-square}      8: 7  J 3  J 5  J 17  J 23
24 = .5^7 . mbox{big non-square}      8: 5  J 7  J 13  J 17  J 23
25 =  . mbox{big non-square}      8: 1  J 3  J 7  J 37  J 47
26 = .7^7.53 . mbox{big non-square}      8: 3  J 11  J 17  J 19  J 37
27 =  . mbox{big non-square}      8: 1  J 13  J 31  J 43  J 47
28 = .29 . mbox{big non-square}      8: 5  J 3  J 11  J 13  J 17
29 = .59 . mbox{big non-square}      8: 3  J 3  J 7  J 19  J 31
30 = .31.61 . mbox{big non-square}      8: 3  J 11  J 19  J 23  J 47
31 = .43 . mbox{big non-square}      8: 3  J 5  J 11  J 13  J 23
32 =  . mbox{big non-square}      8: 5  J 13  J 37  J 47  J 59
33 = .67 . mbox{big non-square}      8: 7  J 5  J 19  J 23  J 53
34 = .59 . mbox{big non-square}      8: 7  J 3  J 7  J 17  J 19
35 = .71 . mbox{big non-square}      8: 3  J 3  J 7  J 19  J 41
36 = .37.41.73 . mbox{big non-square}      8: 1  J 3  J 7  J 23  J 29
37 =  . mbox{big non-square}      8: 5  J 7  J 13  J 17  J 43
38 =  . mbox{big non-square}      8: 5  J 7  J 13  J 17  J 53
39 = .79 . mbox{big non-square}      8: 3  J 7  J 11  J 19  J 23
40 = .41 . mbox{big non-square}      8: 5  J 3  J 11  J 13  J 19
41 = .83 . mbox{big non-square}      8: 7  J 3  J 5  J 11  J 13
42 = .7^11.43 . mbox{big non-square}      8: 1  J 13  J 29  J 31  J 37
43 = .7^11 . mbox{big non-square}      8: 3  J 5  J 11  J 13  J 23
44 = .7^11.89 . mbox{big non-square}      8: 3  J 3  J 5  J 7  J 19
45 = .7^11 . mbox{big non-square}      8: 3  J 5  J 13  J 23  J 41
46 = .7^11.47 . mbox{big non-square}      8: 5  J 3  J 7  J 29  J 43
47 = .7^11 . mbox{big non-square}      8: 3  J 5  J 13  J 41  J 43
48 = .7^11.97 . mbox{big non-square}      8: 7  J 7  J 17  J 19  J 23
49 =  . mbox{big non-square}      8: 1  J 11  J 17  J 23  J 59
50 = .5^21.101 . mbox{big non-square}      8: 1  J 13  J 29  J 37  J 41
51 = .5^21.103 . mbox{big non-square}      8: 3  J 3  J 5  J 11  J 13
52 = .5^21.53.59 . mbox{big non-square}      8: 3  J 3  J 5  J 11  J 17
53 = .5^21.107 . mbox{big non-square}      8: 7  J 5  J 7  J 13  J 19
54 = .5^21.109 . mbox{big non-square}      8: 1  J 7  J 19  J 23  J 41
55 = .5^23.11^9 . mbox{big non-square}      8: 7  J 3  J 7  J 11  J 23
56 = .5^23.11^9.113 . mbox{big non-square}      8: 7  J 5  J 7  J 11  J 17
57 = .5^23.11^9 . mbox{big non-square}      8: 7  J 3  J 7  J 11  J 23
58 = .5^23.11^9.59 . mbox{big non-square}      8: 5  J 11  J 13  J 19  J 23
59 = .5^23.11^9 . mbox{big non-square}      8: 7  J 3  J 7  J 11  J 23
60 = .5^25.11^9.61 . mbox{big non-square}      8: 3  J 3  J 17  J 41  J 61
61 = .5^25.11^9 . mbox{big non-square}      8: 7  J 3  J 7  J 11  J 23
62 = .5^25.11^9 . mbox{big non-square}      8: 7  J 3  J 7  J 11  J 23
63 = .5^25.11^9.127 . mbox{big non-square}      8: 1  J 3  J 5  J 7  J 19
64 = .5^25.11^9 . mbox{big non-square}      8: 3  J 3  J 7  J 11  J 23
65 = .5^27.11^9.131 . mbox{big non-square}      8: 1  J 17  J 23  J 31  J 41
66 = .5^27.67 . mbox{big non-square}      8: 3  J 5  J 7  J 17  J 19
67 = .5^27.107 . mbox{big non-square}      8: 3  J 3  J 5  J 7  J 11
68 = .5^27.137 . mbox{big non-square}      8: 1  J 3  J 5  J 7  J 17
69 = .5^27.139 . mbox{big non-square}      8: 3  J 11  J 13  J 23  J 31
70 = .5^29.71 . mbox{big non-square}      8: 7  J 3  J 7  J 19  J 23
71 = .5^29 . mbox{big non-square}      8: 1  J 7  J 13  J 17  J 23
72 = .5^29.73 . mbox{big non-square}      8: 1  J 5  J 11  J 23  J 29
73 = .5^29 . mbox{big non-square}      8: 1  J 7  J 13  J 17  J 23
74 = .5^29.149 . mbox{big non-square}      8: 5  J 3  J 7  J 11  J 17

Alrighty then. The vast majority of $n$ give non-squares because of one of just a few mechanisms. (A) Often the numerator is divisible by a prime $p \leq 2n+1$ to an odd power. (B) After dividing out by all such primes, the quotient is $3,5,7 \bmod 8.$ (C) The quotient is $2 \bmod 3.$ (D) The quotient is a quadratic nonresidue $\bmod 7.$

Indeed, those four conditions rule out all but nine values of $n \leq 12235,$ which are taken care of by the remaining quotient being a quadratic nonresidue $\bmod {17}.$

49 =  . mbox{big non-square}      8: 1  J 11  J 17  J 23  J 59
244 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 47
4082 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 29
4084 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 29
4086 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 29
4088 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 29
4091 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 29
4093 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 29
4094 =  . mbox{big non-square}      8: 1  J 17  J 19  J 23  J 29
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  • 2
    $\begingroup$ but the computer does not care - Yeah, I know $\ldots$ They are cold, heartless machines $\ldots$ :-$($ $\endgroup$
    – Lucian
    Oct 31, 2014 at 16:49
  • 1
    $\begingroup$ @Lucian, the real reason to do it that way was to keep it a recursion in two variables, all integers. $\endgroup$
    – Will Jagy
    Oct 31, 2014 at 16:55
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The first problem, which essentially asks whether $\sqrt{H_n^{(2)}}$, where $H_n^{(2)}$ is a Generalized Harmonic Number, can ever be rational for $n\ge4$, is very hard. I don't see a way to approach it.


The Second Problem

Let $k_n\in\mathbb{Z}$ be so that $2^{k_n}\le n\lt2^{k_n+1}$. The only integer not exceeding $n$ divisible by $2^{k_n}$ is $2^{k_n}$.

Let $d_n=b_n2^{k_n}$ be the LCM of $\{k:1\le k\le n\}$, where $b_n$ is odd. $\dfrac{d_n}{2^{k_n}}$ is the only odd number in $$ \left\{\dfrac{d_n}{k}:1\le k\le n\right\} $$ Therefore, $$ d_n\sum_{k=1}^n\frac1k $$ is an odd number. Since $d_n$ is an even number for $n\ge2$, we get that $$ \sum_{k=1}^n\frac1k $$ cannot be an integer for $n\ge2$.

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  • $\begingroup$ Do you see it? :-) $\endgroup$ Nov 3, 2014 at 20:53
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I've verified there are only 2 cases for the first 100000 partial sums.

Here's the Mathematica code I used:

Count[
  ParallelTable[Sqrt[HarmonicNumber[r, 2]] ∈ Rationals, {r, 100000}],
  True
]

(* 2 *)
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