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So I came across this question

Are there infinitely many primes of the form $ 4n^2+1$

And if we are scavenging for primes we can easily throw the odd $n$'s out of the picture, and we have even values of n, in which case the number we get if of the form $4m+1$ and we can easily modify the euclidean proof for infinitely many primes to say that there are infinitely many primes of the form $4m+1$ and so we can have infinitely many primes of the required form.

But I feel like there's some subtle mistake I'm making here.

Bill Dubuque
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Anuj Jha
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  • Could you elaborate on your cases (in the question)? – whoisit Dec 10 '23 at 05:32
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Dec 10 '23 at 05:35
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    See problem #$4$ of Landau's problems. Since for $n\gt 1$, we have that $n^2+1$ can only possibly be prime if $n$ is even, say $n=2m$, we get $4m^2+1$, i.e., what you're asking about. As stated in that Wikipedia article, this problem is still currently unsolved. Also, FYI, see the related Edmund Landau's Problems question and answer here. – John Omielan Dec 10 '23 at 06:32
  • parfor(x=1,10000,4*x^2+1,c,if(isprime(c),print1(x","))) proves useful in PARI/gp – Roddy MacPhee Dec 10 '23 at 21:47
  • @BillDubuque Sorry if I'm wrong but there is like 1 implication in my entire argument as pointed out by GregMartin and so I'm confused as to how to make it more obvious. – Anuj Jha Dec 11 '23 at 06:13
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    "..throw odd $n$'s out of the picture..." What does that statement mean? Because $4\cdot 3^2+1=37$ and $4\cdot5^2+1=101$, etc. – Keith Backman Dec 11 '23 at 16:47

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You've incorrectly used an implication in the wrong direction. The fact that every number of the form $4n^2+1$ is also of the form $4m+1$ means that (infinitely many primes of the form $4n^2+1$) implies (infinitely many primes of the form $4m+1$); but you used the converse of this implication. (Replace $4n^2+1$ with $(4n+1)^2$ to see how this error can easily prove a false statement!)

[By the way, if you haven't yet seen the Euclidean-style proof for the fact that there are infinitely many primes of the form $4m+1$: it's not as straightforward as you might think (it's harder than the corresponding proof for $4m+3$).]

Greg Martin
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  • Ah thanks!

    For the proof I did consult a MSE post later https://math.stackexchange.com/q/244915/1047063

     – Anuj Jha Dec 11 '23 at 06:14