I see it in a popular science book, but there is not any proof. And I try some numbers, for example, $T=1,3$. Since I really have not any knowledge of number theory. So, I asked here. Maybe, it is such easy that it is not suitable asked. If so, after I know it, I will delete it. Thanks.
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Just expand $(2n+1)^2$ and watch almost everything disappear mod $4$. – Randall Oct 18 '21 at 12:02
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If $T=2n+1$, what’s $-T^2-3$? – J. W. Tanner Oct 18 '21 at 12:02
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The proof *is* just to try for $T=1$ and $T=3$, since those are the only odd numbers (mod 4). – Joe Oct 18 '21 at 12:03
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Thanks all, I really be not clever,,, – Enhao Lan Oct 18 '21 at 12:07
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2It's not about "being clever" (fixed mindset). You admittedly don't have knowledge of number theory, and I'm sure the commenters all do. If you study (growth midset) number theory, you'll know these things (and more). – Joe Oct 18 '21 at 12:13
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2See this post. – Dietrich Burde Oct 18 '21 at 18:20
1 Answers
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If $T=2n+1$, then $T^2= 4n^2+4n+1$. Working mod $4$, this is $1$, but $-1 \equiv 3 \bmod 4$.
Randall
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