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Will Jagy
Will Jagy
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Just to get things started:

It happens that I already did the case of 7 in my original question. There are infinitely many solutions to $ u^2 - 7 v^2 = 2$ in integers, beginning with $ u = 3, \; v = 1.$ For any such pair, the positive binary form $ x^2 + 7 y^2 $ integrally represents the consecutive triple $$ 7 v^2, \; 1 + 7 v^2, \; 2 + 7 v^2 = u^2. $$ For the first and third numbers prime factorization is evident. For the middle number, and indeed anything integrally represented by $ x^2 + 7 y^2,$ we know that for any prime factor $p$ with $(-7 | p) = -1$ the exponent must be even. In this particular case those exponents must be $0$ because of the $1.$ So there are an infinite number of these triples. Things get rapidly more difficult when replacing $7$ by any of $ 71, \; \; 311, \; \; 479, \; \; 1559 $$23, \; \; 71, \; \; 311, \; \; 479, \; \; 1559 $ and asking for longer "legal" intervals.

Just to get things started:

It happens that I already did the case of 7 in my original question. There are infinitely many solutions to $ u^2 - 7 v^2 = 2$ in integers, beginning with $ u = 3, \; v = 1.$ For any such pair, the positive binary form $ x^2 + 7 y^2 $ integrally represents the consecutive triple $$ 7 v^2, \; 1 + 7 v^2, \; 2 + 7 v^2 = u^2. $$ For the first and third numbers prime factorization is evident. For the middle number, and indeed anything integrally represented by $ x^2 + 7 y^2,$ we know that for any prime factor $p$ with $(-7 | p) = -1$ the exponent must be even. In this particular case those exponents must be $0$ because of the $1.$ So there are an infinite number of these triples. Things get rapidly more difficult when replacing $7$ by any of $ 71, \; \; 311, \; \; 479, \; \; 1559 $ and asking for longer "legal" intervals.

Just to get things started:

It happens that I already did the case of 7 in my original question. There are infinitely many solutions to $ u^2 - 7 v^2 = 2$ in integers, beginning with $ u = 3, \; v = 1.$ For any such pair, the positive binary form $ x^2 + 7 y^2 $ integrally represents the consecutive triple $$ 7 v^2, \; 1 + 7 v^2, \; 2 + 7 v^2 = u^2. $$ For the first and third numbers prime factorization is evident. For the middle number, and indeed anything integrally represented by $ x^2 + 7 y^2,$ we know that for any prime factor $p$ with $(-7 | p) = -1$ the exponent must be even. In this particular case those exponents must be $0$ because of the $1.$ So there are an infinite number of these triples. Things get rapidly more difficult when replacing $7$ by any of $23, \; \; 71, \; \; 311, \; \; 479, \; \; 1559 $ and asking for longer "legal" intervals.

Source Link
Will Jagy
Will Jagy
  • 25.7k
  • 2
  • 65
  • 121

Just to get things started:

It happens that I already did the case of 7 in my original question. There are infinitely many solutions to $ u^2 - 7 v^2 = 2$ in integers, beginning with $ u = 3, \; v = 1.$ For any such pair, the positive binary form $ x^2 + 7 y^2 $ integrally represents the consecutive triple $$ 7 v^2, \; 1 + 7 v^2, \; 2 + 7 v^2 = u^2. $$ For the first and third numbers prime factorization is evident. For the middle number, and indeed anything integrally represented by $ x^2 + 7 y^2,$ we know that for any prime factor $p$ with $(-7 | p) = -1$ the exponent must be even. In this particular case those exponents must be $0$ because of the $1.$ So there are an infinite number of these triples. Things get rapidly more difficult when replacing $7$ by any of $ 71, \; \; 311, \; \; 479, \; \; 1559 $ and asking for longer "legal" intervals.