I saw a previous proof on this site that addresses this question, but the proof leaves off with a number of form $4k-1$ can't divide $t^2+1$. Why is this? Thanks.
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4Can you add a link to the previous proof for reference? – Ian Miller Jan 08 '17 at 15:28
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Are you talking about this question? – Jyrki Lahtonen Jan 08 '17 at 15:36
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I worded my old edit poorly. I have edited the question and the link is:http://math.stackexchange.com/questions/245299/integer-solutions-for-x2-y3-23 – Hostile Amigo Jan 08 '17 at 15:38
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See here concerning solutions of $y^2 = x^3 +k $ but not for $k=23$. – Jan 08 '17 at 15:41
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I was barred from posting my answer by the closure. Anyway, the point is that an integer of the form $4k-1$ has a prime factor $p$ (at least one) of the same form. It follows that $p\nmid t^2+1$ because then $-1$ would be a quadratic residue modulo $p$. But this is possible if and only if $p\equiv1\pmod4$. – Jyrki Lahtonen Jan 08 '17 at 15:45
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This is starting to make sense. But why can't p be congruent to 1 (mod 4)? – Hostile Amigo Jan 08 '17 at 15:46
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OH! p cant be congruent to 1 mod 4 because it is of form 4k-1. is this right? – Hostile Amigo Jan 08 '17 at 15:48
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$p$ can be congruent to $1$, but at least one of the prime factors of $4k-1$ is $\equiv -1\pmod4$. I am using that particular prime, not just any prime factor of $4k-1$. – Jyrki Lahtonen Jan 08 '17 at 15:48
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thank you so much, it all makes sense now – Hostile Amigo Jan 08 '17 at 15:49
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1I think we can leave this closed as a duplicate. Even though this question was about a detail left open in Manzoni's answer, the point was covered in Elkies' answer. – Jyrki Lahtonen Jan 08 '17 at 15:52