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So today we were trying to prove algebraically that two consecutive integers n, n+1 where each is squared mod 4 is 1. We got quite far, but I can't for the life of me find the notes which we made.

Not much more I can write really, if I find my notes I'll updated the question.

To rephrase the question: prove that (n^2 + (n+1)^2) mod 4 = 1

Jacob Garby
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  • Are you sure it's not supposed to be proving that $n^2 + (n+1)^2 \equiv 1 \mod 4$? The question says consecutive squared integers but the text doesn't mention it... – 2012ssohn Nov 10 '16 at 21:51
  • Your title says "sum of any two consecutive squared integers" but in your post you mention only the sum of consecutive integers. Please check your question. – Xam Nov 10 '16 at 21:55
  • @Charter oh deary me you're right, updated question – Jacob Garby Nov 10 '16 at 22:23

1 Answers1

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Hint $\ $ One of $\,n,n\!+\!1\,$ is even, the other is odd, and $ $ even$^2\equiv 0,\ $ odd$^2\!\equiv\{1,3\}^2\equiv 1\pmod 4$

Alternatively $\, n^2\!+(n\!+\!1)^2 = 2n(n\!+\!1)+1\,$ and, again, use one of $\,n,\,n\!+\!1\,$ is even.

Bill Dubuque
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  • could you explain this first option? what does the {1,3}^2 mean? – Jacob Garby Nov 10 '16 at 23:01
  • @Jacob Odd integers have form $,4n+1,$ or $,4n+3,,$ i.e. mod $4$ they are $\equiv 1$ or $3$, both of whose squares are $\equiv 1,,$ i.e. ${1,3}^2 \equiv {1^2,3^2}\equiv {1}.,$ It's clearer if you use $,3\equiv -1,$ then we can write it as odd$^2 \equiv (\pm1)^2\equiv 1\pmod4\ \ $ – Bill Dubuque Nov 10 '16 at 23:56
  • ah yes I see what you mean, thank you :) – Jacob Garby Nov 11 '16 at 08:07