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A232531 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.

%I A232531 #78 Jan 02 2024 15:09:59
%S A232531 3,5,6,10,11,12,13,15,19,20,21,22,24,26,27,29,30,33,35,37,38,39,40,42,
%T A232531 43,44,45,48,51,52,53,54,55,57,58,59,60,61,65,66,67,69,70,74,75,76,77,
%U A232531 78,80,83,84,85,86,87,88,90,91,93,95,96,99,101,102,104,105,106
%N A232531 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.
%C A232531 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
%C A232531 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.
%C A232531 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).
%C A232531 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.
%C A232531 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.
%C A232531 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.
%C A232531 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.
%C A232531 Positive numbers that are not the difference between two legs of a Pythagorean right triangle. - _Michael Somos_, Apr 02 2017
%H A232531 Robert Israel, <a href="/A232531/b232531.txt">Table of n, a(n) for n = 1..10000</a>
%H A232531 V. Raman, <a href="/A232531/a232531.txt">Proof for individual terms</a>
%e A232531 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.
%e A232531 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.
%e A232531 n = 7 is not a member of this sequence because 15 = 1^2 + 14*1^2 = 2*2^2 + 7*1^2.
%p A232531 filter:= n -> [isolve(x^2-2*y^2=n)]=[] and [isolve(x^2-2*y^2=-n)]=[]:
%p A232531 select(filter, [$1..200]); # _Robert Israel_, Apr 29 2020
%o A232531 (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", ")))
%o A232531 (Python)
%o A232531 from itertools import count, islice
%o A232531 from sympy import factorint
%o A232531 def A232531_gen(): # generator of terms
%o A232531     return filter(lambda n:any((2 < p & 7 < 7) and e & 1 for p, e in factorint(n).items()),count(1))
%o A232531 A232531_list = list(islice(A232531_gen(),30)) # _Chai Wah Wu_, Jun 28 2022
%Y A232531 Cf. A035251, A232532, A232681, A232682.
%K A232531 nonn
%O A232531 1,1
%A A232531 _V. Raman_, Nov 25 2013