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A232531
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Numbers n such that the equation a^2 + 2*n*b^2 = 2*c^2 + n*d^2 has no solutions in positive integers for a, b, c, d.
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5
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3, 5, 6, 10, 11, 12, 13, 15, 19, 20, 21, 22, 24, 26, 27, 29, 30, 33, 35, 37, 38, 39, 40, 42, 43, 44, 45, 48, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 65, 66, 67, 69, 70, 74, 75, 76, 77, 78, 80, 83, 84, 85, 86, 87, 88, 90, 91, 93, 95, 96, 99, 101, 102, 104, 105, 106
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OFFSET
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1,1
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COMMENTS
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Numbers n such that the equation |x^2 - 2y^2| = n has no solutions in integers x,y. In other words, this sequence is the complement of A035251. - Thomas Ordowski and Altug Alkan, Feb 10 2017
It appears that this is the set of all numbers which contain at least one prime factor p congruent to 3 (mod 8) or 5 (mod 8) raised to an odd power.
With n = 3, the equation a^2 + 6*b^2 = 2*c^2 + 3*d^2 has no solutions in positive integers for a, b, c, d as the following proof shows: Let's assume that gcd(a, b, c, d) = 1, otherwise if gcd(a, b, c, d) = g, then a/g, b/g, c/g, d/g would be a smaller set of solutions to the equation. Considering modulo 3 arithmetic, we have a^2 - 2*c^2 == 0 (mod 3). Since a square is always congruent to 0 (mod 3) or 1 (mod 3), this is possible if and only if a == 0 (mod 3) and c == 0 (mod 3). Now let a = 3*p, c = 3*q, so a^2 = 9*p^2, c^2 = 9*q^2. Substituting this into the equation a^2 + 6*b^2 = 2*c^2 + 3*d^2 gives 9*p^2 + 6*b^2 = 18*q^2 + 3*d^2, i.e. 3*p^2 + 2*b^2 = 6*q^2 + d^2. Taking modulo 3 arithmetic with this equation again gives 2*b^2 - d^2 == 0 (mod 3). By using the same argument as above, this is possible if and only if b == 0 (mod 3) and d == 0 (mod 3). We already showed that a == 0 (mod 3) and c == 0 (mod 3), so gcd(a, b, c, d) should be a multiple of 3. This contradicts our assumption that gcd(a, b, c, d) = 1 and a/3, b/3, c/3, d/3 are a smaller set of solutions to the above mentioned equation. By using the proof of infinite descent, this implies that the only possible set of solutions to (a, b, c, d) is (0, 0, 0, 0).
We can similarly prove for the other values of n by taking modulo n arithmetic if the only solution to a^2 - 2*d^2 == 0 (mod n) is a == 0 (mod n) and d == 0 (mod n). If r is a prime factor of n and if r^2 does not divide n and the equation a^2 - 2*d^2 == 0 (mod r) has the only solution a == 0 (mod r) and d == 0 (mod r), we can also take modulo r arithmetic to prove that n is a member of this sequence. If n is odd, n is a member of this sequence if and only if 2 is a quadratic non-residue (mod n). Alternately, n should have at least one prime factor congruent to 3 (mod 8) or 5 (mod 8) raised to an odd power.
If n = 2*k is even and not a multiple of 4, taking modulo 2 arithmetic yields a to be even. Putting a = 2*p, and dividing the equation by 2 gives 2*(p^2+k*b^2) = (c^2+k*d^2). This equation will have no solution in positive integers p, b, c, d if and only if there is no number that can be written by the form x^2+k*y^2 that is twice another number that can be written by the same form x^2+k*y^2. If n is even, n is a member of this sequence if and only if n has at least one prime factor congruent to 3 (mod 8) or 5 (mod 8) raised to an odd power.
If n is a multiple of 4, then n = 4*m is a member of this sequence if and only if m is a member of this sequence.
Also numbers n such that there is no number that can be written by the form x^2+n*y^2 that is twice another number that can be written by the same form x^2+n*y^2.
Positive numbers that are not the difference between two legs of a Pythagorean right triangle. - Michael Somos, Apr 02 2017
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LINKS
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EXAMPLE
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n = 3 is a member of this sequence because there is no positive integer m which can be simultaneously written as both x^2+6*y^2 and 2*x^2+3*y^2. The former requires the sum of {2, 3, 5, 11} mod 24 prime factors of m to be even, while the latter requires the sum of {2, 3, 5, 11} mod 24 prime factors of m to be odd.
n = 5 is a member of this sequence because there is no positive integer m which can be simultaneously written as both x^2+10*y^2 and 2*x^2+5*y^2. The former requires the sum of {2, 5, 7, 13, 23, 37} mod 40 prime factors of m to be even, while the latter requires the sum of {2, 5, 7, 13, 23, 37} mod 40 prime factors of m to be odd.
n = 7 is not a member of this sequence because 15 = 1^2 + 14*1^2 = 2*2^2 + 7*1^2.
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MAPLE
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filter:= n -> [isolve(x^2-2*y^2=n)]=[] and [isolve(x^2-2*y^2=-n)]=[]:
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PROG
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(PARI) for(n=1, 10000, flag=0; v=factor(n); for(i=1, matsize(v)[1], if((v[i, 1]%8==3||v[i, 1]%8==5)&&v[i, 2]%2==1, flag=1; break)); if(flag==1, print1(n", ")))
(Python)
from itertools import count, islice
from sympy import factorint
def A232531_gen(): # generator of terms
return filter(lambda n:any((2 < p & 7 < 7) and e & 1 for p, e in factorint(n).items()), count(1))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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