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.
Show that $y^2=x^3+23$ has no integer solutions (hint: use $y^2+4$)
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elementary-number-theory
diophantine-equations
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4Can you add a link to the previous proof for reference? – 2017-01-08
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0Are you talking about [this question](http://math.stackexchange.com/q/245299/11619)? – 2017-01-08
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0I 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 – 2017-01-08
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0See [here](https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf&ved=0ahUKEwi2696P8LLRAhUEso8KHam3DogQFggjMAM&usg=AFQjCNExJsQ8bikneg2KeVVzvOH_vwlctQ&sig2=0YsUxMNskPYIBFdvu2aSOQ) concerning solutions of $y^2 = x^3 +k $ but not for $k=23$. – 2017-01-08
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0I 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 $p\equiv1\pmod4$. – 2017-01-08
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0This is starting to make sense. But why can't p be congruent to 1 (mod 4)? – 2017-01-08
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0OH! p cant be congruent to 1 mod 4 because it is of form 4k-1. is this right? – 2017-01-08
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0$p$ 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 $4k-1$. – 2017-01-08
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0thank you so much, it all makes sense now – 2017-01-08
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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](http://math.stackexchange.com/a/579939/11619). – 2017-01-08