Powerful Values of Quadratic Polynomials

Jean-Marie De Koninck and Nicolas Doyon Départment de Mathématiques Université Laval Québec, PQ G1V 0A6 Canada

                      jmdk@mat.ulaval.ca
                   nicolas.doyon.1@ulaval.ca

Florian Luca Instituto de Matemáticas Universidad Nacional Autónoma de México C. P. 58180 Morelia, Michoacán México

                     fluca@matmor.unam.mx

Abstract We study the set of those integers k such that n^{2} + k is powerful for infinitely many

positive integers n. We prove that most integers k have this property.

1 Introduction

Given an arbitrary integer k ≠ 0, Mollin and Walsh [1] have shown that there exist infinitely many ways of writing k as a difference of two nonsquare powerful numbers. A positive integer n is said to be powerful if p^{2} | n for each prime divisor p of n. For instance, 17 has two such representations below 10^{9}, namely 17 = 125 − 108 and 17 = 173881809 − 173881792. But what about representations 17 = P − n^{2}, where P is a powerful number? It turns out that there are no such representation with P < 10^{9}. However, in view of Theorem 1 (below) we believe that infinitely many such representations should exist, even though the smallest is probably very large (see Table 1 in Section 3). In general, identifying all those integers k ≠ 0 such that n^{2} + k is a powerful number for infinitely many positive integers n seems to be a very difficult problem. Indeed, already showing that one such n exists is not obvious.

Now, for a given k ≠ 0, if one can find an integer n_{0} such that n_{0}^{2} +k is a powerful number which is not a perfect square, that is if

n_{0}^{2} + k = m_{0}^{2}b^{3},  (1)

for some integer m_{0}, with b > 1 squarefree, then (1) can be written as

n_{0}^{2} − Dm_{0}^{2} = −k,  (2)

where D > 1 is a cube but not a square. Now since x^{2} − Dy^{2} = −k is a generalized Pell equation with a given solution (see, for instance, Robertson [4]), it must have infinitely many solutions, thus providing infinitely many n’s for which n^{2} + k is powerful. However, given an arbitrary integer k, finding a minimal solution (n_{0}, m_{0}) of (2) for an appropriate D is not easily achieved.

In this note, we use a different approach and show that for almost all integers k, there exist infinitely many positive integers n such that n^{2} + k is powerful. In fact, we prove the following result.

Theorem 1. For any positive real number x, let P(x) be the set of integers k with |k| ≤ x such that n^{2} +k is powerful for infinitely many positive integers n. Then 2x−#P(x) = o(x).

Throughout this paper we use the Landau symbols O and o as well as the Vinogradov symbols ≫ and ≪ with their usual meaning. We let log_{1} x = max{1, log x}, where log stands for the natural logarithm, and for i > 1 we define log_{i} x = log_{1} (log_{i−1} x). When i = 1 we omit the subscript and thus understand that all the logarithms that will appear are ≥ 1. For a positive integer n we write ϕ(n) for the Euler function of n.

2 The Proof

For the proof, we first give a sufficient algebraic criterion on k which insures that n^{2} + k is powerful for infinitely many n. We then show that most integers k satisfy this condition.

We shall prove this only when k > 0, but the argument extends without any major modification to the case k < 0.

Proposition 2. Assume that there exist positive integers y_{1} and d | ky_{1}^{2} + 1 such that

u = \frac{1}{2y_{1}} (\frac{ky_{1}^{2} + 1}{d} − d)

is a positive integer coprime to k. Let D = u^{2} + k and assume further that y_{1} is coprime to D. Then there exist infinitely many positive integers n such that n^{2} + k is powerful.

Proof. Since u is an integer, it follows that d and (ky_{1}^{2} + 1)/d are integers of the same parity. Put

x_{1} = \frac{1}{2} (\frac{ky_{1}^{2} + 1}{d} + d) .

One checks immediately that x_{1}^{2} − (uy_{1})^{2} = ky_{1}^{2} + 1, which can be rewritten as

x_{1}^{2} − Dy_{1}^{2} = 1.  (3)

Define the sequences (x_{m})_{m≥1} and (y_{m})_{m≥1} as

x_{m} + y_{m}\sqrt{D} = (x_{1} + y_{1}\sqrt{D})^{m}

for all m ≥ 1. Then, for all m ≥ 1,

x_{m}^{2} − Dy_{m}^{2} = (x_{m} + y_{m}\sqrt{D})(x_{m} − y_{m}\sqrt{D})  (4)
= (x_{1} + y_{1}\sqrt{D})^{m}(x_{1} − y_{1}\sqrt{D})^{m}
= (x_{1}^{2} − Dy_{1}^{2})^{m} = 1

We now search for positive integers n such that n^{2} + k = Dℓ^{2} holds with some positive integer ℓ such that D | ℓ. It is clear that such numbers n have the property that n^{2} + k is powerful. We rewrite this equation as

n^{2} − Dℓ^{2} = −k.  (5)

Noting that u^{2} − D · 1^{2} = −k and using (4), if

n + \sqrt{D}ℓ = (u + \sqrt{D})(x_{m} + \sqrt{D}y_{m}),

one checks by multiplying each side by its conjugate that the pair (n, ℓ) satisfies (5). Expanding we get

n = ux_{m} + Dy_{m} and ℓ = uy_{m} + x_{m}.

It suffices to argue that there exist infinitely many m such that D | ℓ. Since

x_{m} = \frac{1}{2} ((x_{1} + y_{1}\sqrt{D})^{m} + (x_{1} − y_{1}\sqrt{D})^{m}) ≡ x_{1}^{m} (mod D),

and

y_{m} = \frac{1}{2\sqrt{D}} ((x_{1} + y_{1}\sqrt{D})^{m} − (x_{1} − y_{1}\sqrt{D})^{m}) ≡ mx_{1}^{m−1}y_{1} (mod D),

the relation D | (uy_{m} +x_{m}) is equivalent to D | x_{1}^{m−1}(umy_{1} +x_{1}). Since D and x_{1} are coprime (in light of (3)), the above divisibility relation holds if and only if muy_{1} ≡ −x_{1} (mod D). Since both u and y_{1} are coprime to D, it follows that their product is invertible modulo D. Hence, if m ≡ −x_{1}(uy_{1})^{−1} (mod D), then

n = ux_{m} + Dy_{m}

has the property that n^{2} + k is powerful. This completes the proof of the proposition. ⊓ ⊔

It now remains to show that for most positive integers k one can choose integers y_{1} and d such that the conditions from Proposition 2 are fulfilled. It is clear that for the purpose of making n^{2} + k powerful, we may assume that k is squarefree. Indeed, if p^{2} | k, we may then take n = pn′ and note that

n^{2} + k = p^{2}(n^{′2} + (k/p^{2})),

so we may replace k by k/p^{2}.

Theorem 3. The set of squarefree positive integers k for which there exist positive integers y_{1} and d | ky_{1}^{2} + 1 such that the conditions of Proposition 2 are satisfied is of density 1.

Proof. We let x be a large positive real number and we assume that k ≤ x is a positive integer. We choose y_{1} = 12.

The number d will always be a prime number in a certain arithmetical progression modulo 144, as follows. If gcd(k, 6) = 1, we then take d ≡ 1 (mod 144). If gcd(k, 6) = 2, then d ≡ 91 (mod 144). If gcd(k, 6) = 3, we put d ≡ 65 (mod 144), and finally if 6 | k, we then put d ≡ 11 (mod 144).

We first show that y_{1} and D are coprime, from which it will follow that 6 and D are coprime.

If gcd(k, 6) = 1, then since d ≡ 1 (mod 144), we get that (144k + 1)/d ≡ 1 (mod 144). Hence, (144k + 1)/d − d ≡ 0 (mod 144), which shows that 6 | u. Since k is coprime to 6, we get that 6 is coprime to D.

If gcd(k, 6) = 2, then d ≡ 91 (mod 144). In particular, d ≡ 11 (mod 16) and d ≡ 1 (mod 9). Hence, (144k + 1)/d ≡ 3 (mod 16) and (144k + 1)/d ≡ 1 (mod 9). Thus, (144k +1)/d−d is congruent to 8 modulo 16 and to 0 modulo 9. Hence, u is an odd multiple of 3. Since 2 divides k but 3 doesn’t, we get that 6 is coprime to D.

It is easily seen that the other two cases, namely gcd(k, 6) = 3 and gcd(k, 6) = 6, can be treated similarly.

Moreover, since gcd(u^{2} + k, 12) = gcd(D, y) = 1 there is no prime p ∈ {2, 3} such that p | gcd(k, u).

Thus, it remains to show that for all positive integers k ≤ x except o(x) of them such a prime d can be chosen in such a way that there is no prime p > 3 dividing both u and k. Note that if p > 3 divides both u and k, then (144k + 1)/d ≡ d (mod p), so that 144k + 1 ≡ d^{2} (mod p), in which case d^{2} ≡ 1 (mod p). Thus, d ≡ ±1 (mod p). We can reverse the argument to show that if d ≡ ±1 (mod p), then p | 2y_{1}u and p | k. Since p > 3 and the largest prime factor of y_{1} is 3, the condition d ≡ ±1 (mod p) guarantees that p | gcd(u, k).

For coprime positive integers a, b we write

S(x; a, b) = ∑_{p≡a (mod b)} \frac{1}{p} − \frac{log_{2} x}{ϕ(b)}.

A result of Pomerance (see Theorem 1 and Remark 1 in [3]) shows that, uniformly for all a < b ≤ x,

S(x; a, b) = \frac{1}{p(a, b)} + O (\frac{log 2b}{ϕ(b)}) ,

where p(a, b) is the smallest prime number in the arithmetical progression a (mod b).

Let ω(k; a, b) be the number of prime factors p of k which are congruent to a (mod b). We let b = 144, a = 1, 91, 65, 11 according to whether gcd(k, 6) = 1, 2, 3, 6, respectively. By a classical result of Turán [5], the estimate

ω(144k + 1; a, b) = \frac{log_{2} x}{ϕ(b)} + O ((\frac{log_{2} x}{ϕ(b)})^{2/3})

holds for all k ≤ x, with at most

O (\frac{x}{(log_{2} x)^{1/6}} ) = o(x)

exceptions. From now on, we work only with such positive integers k. Note that ω(144k + 1; a, b) gives the number of admissible values for d.

We now put y = log_{2} x, z = x^{1/3} and show that the number of such k ≤ x for which there exists a prime factor p ∈ [y, z] dividing both u and k is o(x). Indeed, let us fix p and d. Then k ≡ 0 (mod p) and 144k + 1 ≡ 0 (mod d). This puts k ≤ x into a certain arithmetical progression modulo pd.

Assume first that pd ≤ x. Clearly, the number of such positive integers k ≤ x is ≤ x/(pd) + 1 ≪ x/(pd). Summing up this inequality for all p ≥ y and all d ≡ ±1 (mod p), we get that the number of such numbers k is

≪ x ∑_{p≥y} \frac{1}{p} ∑_{\begin{matrix}d≤x \\ d ≡ ±1 (mod p) \end{matrix}} \frac{1}{d} ≪ x log_{2} x ∑_{p≥y} \frac{1}{p^{2}} ≪ \frac{x log_{2} x}{y log y} = o(x).

We now look at those positive integers k ≤ x such that pd > x. Write 144k+1 = dm, and note that m ≤ 288x/d < 288p ≪ p. Since p | k and d ≡ ±1 (mod p), we get that m ≡ ±1 (mod p). Fix m. Then k/p is in a certain residue class modulo m depending on p. Write k/p = v + mℓ. Then

dm = 144p(k/p) + 1 = (144pv + 1) + 144pmℓ,

so that

d = w + 144pℓ,

where w = (144pv + 1)/m. Furthermore, d ≤ 288x/m. Hence, by a result of Montgomery and Vaughan [2], the number of such primes d does not exceed

\frac{4 · 144x}{mϕ(144p) log(288x/(mp))} ≪ \frac{x}{mp log x},  (6)

where we used the fact that m ≤ 288p ≤ 288x^{1/3}, and therefore

\frac{288x}{mp} ≫ x^{1/3}.

Summing up inequality (6) over all the possible values of m ≤ 288p ≤ 288x^{1/3}, and then afterwards over all p ∈ [y, z], we get that the number of such k’s is

≪ \frac{x}{log x} ∑_{y≤p} \frac{1}{p} ∑_{\begin{matrix}m≤x \\ m ≡ ±1 (mod p) \end{matrix}} \frac{1}{m} ≪ x ∑_{y≤p} \frac{1}{p^{2}} ≪ \frac{x}{y log y} = o(x).

From now on, we consider only those k such that if d ≡ a (mod b) is a prime factor of 144k + 1, with the pair (a, b) being the appropriate one depending on the value of gcd(k, 6), then there exists a prime p ∈ [5, y] ∪ [x^{1/3}, x] such that p | gcd(k, u). First observe that k has at most 3 prime factors in [x^{1/3}, x].

Moreover, for each prime p > x^{1/3}, there are at most 3 values of d such that d ≡ ±1 (mod p). Indeed, if there were 4 or more, let d_{1}, d_{2}, d_{3} and d_{4} be 4 of them. We would then have

144x + 1 ≥ 144k + 1 ≥ d_{1}d_{2}d_{3}d_{4} ≥ (x^{1/3} − 1)^{4},

which is impossible for large x. Hence, there are at most 9 values of d which might be

congruent to ±1 modulo some prime factor p > x^{1/3} of k. Since we have (1 + o(1))\frac{log_{2} x}{ϕ(b)}

such primes d, it follows that we also have (1 + o(1))\frac{log_{2} x}{ϕ(b)} such prime factors d of 144k + 1

with the property that each of them is congruent to ±1 modulo a prime factor p of k in the interval [5, y]. We apply Turán’s inequality from [5] again to conclude that all k ≤ x have at most 1.5 log_{2} y < 2 log_{4} x prime factors p < y with at most o(x) exceptions.

We now write

M = ∏_{5≤p<(log_{3} x)/2} p,

and look at those d such that d ≡ 2 (mod M). Note that such d are in a certain arithmetical progression A (mod B), where B = bM = (log_{2} x)^{1/2+o(1)}. We apply again the results from [3] and [5] to infer that all positive integers k ≤ x have ω(k; A, B) factors in the interval

[\frac{log_{2} x}{2ϕ(B)}, \frac{2 log_{2} x}{ϕ(B)}] ,

with o(x) possible exceptions. Because d ≡ 2 (mod M), we have that d ≢ ±1 (mod p) for all p < (log_{3} x)/2. Hence, there exist at least

μ := ⌊(log_{2} x)/(4 log_{4} xϕ(B))⌋ > (log_{2} x)^{1/3}

such primes d which furthermore are congruent to either 1 or −1 modulo p for the same prime p > (log_{3} x)/2.

We now count how many such k’s can there can be. Because of the above argument, we can write 144k + 1 = d_{1}d_{2} ⋯ d_{μ}Q < 288x (for some positive integer Q), where each d_{j} ≡ ±1 (mod p). Thus, the number of such k’s is at most

∑_{p>(log_{3} x)/2} \frac{288x}{μ!} ( ∑_{\begin{matrix}d≤x \\ d ≡ A (mod B) \\ d ≡ ±1 (mod p) \end{matrix}} \frac{1}{d})^{μ} ≪ x ∑_{p>(log_{3} x)/2} (\frac{2e log_{2} x + O(1)}{μϕ(B)(p − 1)})^{μ}

≪ x ∑_{p>(log_{3} x)/2} (\frac{O(log_{4} x)}{p})^{(log_{2} x)^{1/3}}
= o(x),

which completes the proof of Theorem 3. ⊓ ⊔

3 Comments and Numerical Results

Although we proved that n^{2} + k is powerful for infinitely many n’s only for most integers k, we do conjecture that this is actually true for all integers k. Indeed, fixing a squarefree integer k, the probability that n^{2} + k is powerful is of the order \frac{1}{\sqrt{n^{2}+k}} ≈ \frac{1}{n} for large n. This means that we should expect that

#\{n ≤ x : n^{2} + k is powerful\} ∼ ∑_{n≤x} \frac{1}{n} ∼ log x → ∞ as x → ∞  (7)

for any squarefree k. From our proof it follows indeed that if #{n : n^{2} + k is powerful} > 1, then #{n ≤ x : n^{2} + k is powerful} ≫ log x.

Table 1 (resp. Table 2) provides, for each integer 1 ≤ k ≤ 50, the smallest known value of n for which n^{2} +k (resp. n^{2} −k) is a powerful number without being a perfect square. These values of n were obtained by finding the minimal solution of x^{2} − Dy^{2} = ±k by considering various cubefull D’s. Those n > 10^{9} may not be the smallest n for which n^{2} ± k is powerful.

Given three integers a, b, c, one could ask if the polynomial an^{2} + bn + c is powerful for infinitely many integers n. Assuming that an^{2} + bn + c is a powerful number which is not a square, we can then write an^{2} + bn + c = Dm^{2} with D > 1 squarefree and D | m. We then have

n = \frac{−b ± \sqrt{4aDm^{2} + b^{2} − 4ac}}{2a}.

Since n is an integer, there exists an integer y such that 4aDm^{2} + b^{2} − 4ac = y^{2}, or, equivalently, y^{2} − aD(2m)^{2} = b^{2} − 4ac with y ≡ ±b (mod 2a). But then the existence of one solution implies the existence of infinitely many. On the other hand, we also get that if there is an infinity of integers y for which y^{2} − b^{2} + 4ac = Dx^{2} where D > 1 is squarefree and 2aD | x, then there exist infinitely many n’s for which an^{2} + bn + c is a powerful number.

The prediction (7) may at first seem at odd with the fact that some of the smallest n’s obtained in Tables 1 and 2 are huge. However, the statement “n^{2} + k is powerful” is equivalent to “n^{2} + k = dm^{2} with d squarefree and d | m”. Now for any fixed d, this last equation is a generalized Pell equation. While a solution may not exist for some values of d, when it does for a particular d, it is well known that the smallest solution can be surprisingly large. When solutions exist, it is still possible that none of them will be such that d | m. The size of the smallest solution such that d | m can also be quite large. There are thus three possible reasons for the huge value of the smallest solution: a large value of d, a large value of the smallest solution to n^{2} + k = dm^{2} or a large value of the smallest solution such that d | m. We investigated this in Table 3, where n = n_{0} is the smallest solution to n^{2} + k = dm^{2} not taking into account the restriction d | m. It turns out that the surprisingly large values of the smallest n in Table 1 are not due to a very large value of d but rather to a large value of n_{0} (see for instance k = 33), or to the large size of the smallest solution n_{0} for which d | m (see for instance k = 17).

Table 1

\begin{matrix}k & n & n^{2} + k \\ 1 & 682 & 5^{3} · 61^{2} \\ 2 & 5 & 3^{3} \\ 3 & 37 & 2^{2} · 7^{3} \\ 4 & 11 & 5^{3} \\ 5 & 1879706 & 3^{5} · 7^{2} · 23^{3} · 29^{3} \\ 6 & 463 & 5^{4} · 7^{3} \\ 7 & 11 & 2^{7} \\ 8 & 10 & 2^{2} · 3^{3} \\ 9 & 2046 & 3^{2} · 5^{3} · 61^{2} \\ 10 & 341881 & 11^{3} · 9371^{2} \\ 11 & 31 & 2^{2} · 3^{5} \\ 12 & 74 & 2^{4} · 7^{3} \\ 13 & 70 & 17^{3} \\ 14 & 5519 & 3^{3} · 5^{5} · 19^{2} \\ 15 & 793 & 2^{7} · 17^{3} \\ 16 & 22 & 2^{2} · 5^{3} \\ 17 & n_{17} & f_{17} \\ 18 & 57 & 3^{3} · 11^{2} \\ 19 & 559 & 2^{2} · 5^{7} \\ 20 & 338 & 2^{3} · 3^{3} · 23^{2} \\ 21 & n_{21} & f_{21} \\ 22 & 503259461 & 47^{3} · 491^{2} · 3181^{2} \\ 23 & 45 & 2^{11} \\ 24 & 926 & 2^{2} · 5^{4} · 7^{3} \\ 25 & 190 & 5^{3} · 17^{2} \end{matrix}  \begin{matrix}k & n & n^{2} + k \\ 26 & 109 & 3^{5} · 7^{2} \\ 27 & 36 & 3^{3} · 7^{2} \\ 28 & 62 & 2^{5} · 11^{2} \\ 29 & 436 & 3^{2} · 5^{3} · 13^{2} \\ 30 & 832836278711 & 31^{2} · 59^{2} · 67^{2} · 79^{3} · 9679^{2} \\ 31 & 63 & 2^{5} · 5^{3} \\ 32 & 88 & 2^{5} · 3^{5} \\ 33 & n_{33} & f_{33} \\ 34 & 7037029 & 5^{5} · 7^{5} · 971^{2} \\ 35 & 36 & 11^{3} \\ 36 & 33 & 3^{2} · 5^{3} \\ 37 & n_{37} & f_{37} \\ 38 & 5945 & 3^{2} · 7^{3} · 107^{2} \\ 39 & 31 & 2^{3} · 5^{3} \\ 40 & 52 & 2^{3} · 7^{3} \\ 41 & 78 & 5^{3} · 7^{2} \\ 42 & 720025 & 13^{3} · 31^{3} · 89^{2} \\ 43 & 22364 & 11^{3} · 613^{2} \\ 44 & 62 & 2^{4} · 3^{5} \\ 45 & 96 & 3^{3} · 7^{3} \\ 46 & 50927 & 5^{5} · 11^{2} · 19^{3} \\ 47 & 39 & 2^{5} · 7^{2} \\ 48 & 148 & 2^{6} · 7^{3} \\ 49 & 524 & 5^{3} · 13^{3} \\ 50 & 1325 & 3^{5} · 5^{2} · 17^{2} \end{matrix}

Here, n_{17} = 1952785824219551870 with f_{17} = 3^{2} · 13^{3} · 367^{2} · 7487^{2} · 5054107013^{2}; n_{21} = 4580728614212333152148 with f_{21} = 5^{2} · 31^{2} · 37^{3} · 41611^{2} · 3155673955493^{2}; n_{33} = 2451448196948930 with f_{33} = 7^{2} · 17^{3} · 29^{3} · 31992951041^{2}; n_{37} = 18651116694721032166213875246076 with f_{37} = 317^{3} ·10219159057^{2} ·323370789682598407^{2}.

Table 2

\begin{matrix}k & n & n^{2} − k \\ 1 & 3 & 2^{3} \\ 2 & 11427 & 7^{3} · 617^{2} \\ 3 & n_{3} & f_{3} \\ 4 & 6 & 2^{5} \\ 5 & 73 & 2^{2} · 11^{3} \\ 6 & 62531004125 & 19^{3} · 14831^{2} · 50909^{2} \\ 7 & n_{7} & f_{7} \\ 8 & 20 & 2^{3} · 7^{2} \\ 9 & 15 & 2^{3} · 3^{3} \\ 10 & n_{10} & f_{10} \\ 11 & 56 & 5^{5} \\ 12 & 47 & 13^{3} \\ 13 & 16 & 3^{5} \\ 14 & 33017 & 5^{2} · 11^{3} · 181^{2} \\ 15 & 1138 & 109^{3} \\ 16 & 68 & 2^{9} · 3^{2} \\ 17 & 23 & 2^{9} \\ 18 & 19 & 7^{3} \\ 19 & 762488 & 3^{2} · 5^{3} · 127^{2} · 179^{2} \\ 20 & 146 & 2^{4} · 11^{3} \\ 21 & 1552808 & 43^{3} · 5507^{2} \\ 22 & 47 & 3^{7} \\ 23 & 6234 & 7^{2} · 13^{3} · 19^{2} \\ 24 & 32 & 2^{3} · 5^{3} \\ 25 & 45 & 2^{4} · 5^{3} \end{matrix}  \begin{matrix}k & n & n^{2} − k \\ 26 & 2537 & 23^{5} \\ 27 & 51700 & 13^{3} · 1103^{2} \\ 28 & 54 & 2^{3} · 19^{2} \\ 29 & 426 & 7^{3} · 23^{2} \\ 30 & 83 & 19^{3} \\ 31 & 34 & 3^{2} · 5^{3} \\ 32 & 40 & 2^{5} · 7^{2} \\ 33 & 3601 & 2^{8} · 37^{3} \\ 34 & 948281 & 3^{6} · 47^{3} · 109^{2} \\ 35 & 531783519104 & 29^{3} · 997^{2} · 3415409^{2} \\ 36 & 42 & 2^{6} · 3^{3} \\ 37 & 73 & 2^{2} · 3^{3} · 7^{2} \\ 38 & 16493 & 11^{2} · 131^{3} \\ 39 & n_{39} & f_{39} \\ 40 & 632 & 2^{3} · 3^{3} · 43^{2} \\ 41 & 71 & 2^{3} · 5^{4} \\ 42 & 691888331 & 13^{2} · 79^{3} · 75797^{2} \\ 43 & 5016 & 13^{2} · 53^{3} \\ 44 & 112 & 2^{2} · 5^{5} \\ 45 & 219 & 2^{2} · 3^{2} · 11^{3} \\ 46 & 847 & 3^{3} · 163^{2} \\ 47 & 180190 & 53^{3} · 467^{2} \\ 48 & 94 & 2^{2} · 13^{3} \\ 49 & 56 & 3^{2} · 7^{3} \\ 50 & 57135 & 5^{2} · 7^{3} · 617^{2} \end{matrix}

Here, n_{3} = 15503069909027 with f_{3} = 13^{3} · 239^{2} · 64866401293^{2}; n_{7} = 85227106679780 with f_{7} = 3^{3} · 59^{3} · 36192438539^{2}; n_{10} = 71457130044805582612325294634331 with f_{10} = 3^{3}·13^{3}·43^{2}·6823075915494777091540353511^{2}; n_{39} = 82716851195974 with f_{39} = 7^{2} · 373^{3} · 29287^{2} · 56009^{2}.

4 Acknowledgments

The first author was partially supported by a grant from NSERC, while the third author was supported by projects SEP-CONACYT 37259-E and 37260-E.

Table 3

\begin{matrix}k & d & n_{0} \\ 1 & 5 & 2 \\ 2 & 3 & 2 \\ 3 & 7 & 2 \\ 4 & 5 & 1 \\ 5 & 3 · 23 · 29 & 26258 \\ 6 & 7 & 1 \\ 7 & 2 & 1 \\ 8 & 3 & 2 \\ 9 & 5 & 6 \\ 10 & 11 & 1 \\ 11 & 3 & 1 \\ 12 & 7 & 4 \\ 13 & 17 & 2 \\ 14 & 15 & 1 \\ 15 & 34 & 11 \\ 16 & 5 & 2 \\ 17 & 13 & 10 \\ 18 & 3 & 3 \\ 19 & 5 & 1 \\ 20 & 6 & 2 \\ 21 & 37 & 4 \\ 22 & 47 & 5 \\ 23 & 2 & 3 \\ 24 & 7 & 2 \\ 25 & 5 & 10 \end{matrix}  \begin{matrix}k & d & n_{0} \\ 26 & 3 & 1 \\ 27 & 3 & 9 \\ 28 & 2 & 2 \\ 29 & 5 & 4 \\ 30 & 79 & 7 \\ 31 & 10 & 3 \\ 32 & 6 & 8 \\ 33 & 17 · 29 & 1310 \\ 34 & 35 & 1 \\ 35 & 11 & 3 \\ 36 & 5 & 3 \\ 37 & 317 & 61016 \\ 38 & 7 & 5 \\ 39 & 10 & 1 \\ 40 & 14 & 4 \\ 41 & 5 & 2 \\ 42 & 31 · 13 & 19 \\ 43 & 11 & 1 \\ 44 & 3 & 2 \\ 45 & 21 & 12 \\ 46 & 95 & 7 \\ 47 & 2 & 5 \\ 48 & 7 & 8 \\ 49 & 65 & 4 \\ 50 & 3 & 5 \end{matrix}

References

[1] R. A. Mollin and P. G. Walsh, Proper differences of nonsquare powerful numbers, C. R. Math. Rep. Acad. Sci. Canada 10 (1988), no. 2, 71–76.

[2] H. L. Montgomery and R. C. Vaughan, The large sieve, Mathematika 20 (1973), 119–134.

[3] C. Pomerance, On the distribution of amicable numbers, J. reine angew. Math., 293/294 (1977), 217–222.

[4] J. P. Robertson, Solving the generalized Pell equation x^{2} − Dy^{2} = N,

   http://www.hometown.aol.com/jpr2718/pell.pdf.

[5] P. Turán, Uber einige Verallgemeinerungen eines Satzes von Hardy und Ramanujan, J. London Math. Soc., 11 (1936), 125–133.

2010 Mathematics Subject Classification: Primary 11A05; Secondary 11N25. Keywords: powerful numbers, generalized Pell equation.

(Concerned with sequence \underline{A001694}.)

Received October 27 2010; revised version received February 22 2011. Published in Journal of Integer Sequences, March 25 2011.

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