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Let $A=\{3,4\}$ be a subset of $S=\{1,2,..., 6\}$. Let $n\in S$. Prove if $\frac{n^2(n+1)^2}{4}$ is even, then $n\in A$.

I know this needs to be a proof by cases, and it should be "Assume $n$ is even. By definition $n=2k$ for $k \in \mathbb{Z}$. From here we note that $\frac{n^2(n+1)^2}{4}=\frac{(2k)^2(2k+1)^2}{4}$." From here I am not finding it to be even or odd when it is simplified. I get $k^2(2k+1)^2$, which is not helpful.

Wng427
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    If the wording of the question is exactly as you stated, since there are only $6$ elements in $S$, you can check their $\frac {n^2(n+1)^2}4$ one by one. – player3236 Oct 06 '20 at 03:51
  • either $n$ or $n+1$ is even, so, if $8$ divides $n^2(n+1)^2$, then $8$ divides $n^2$ or $8$ divides $(n+1)^2$; if $8$ divides $N^2$, then $4$ divides $N$ – J. W. Tanner Oct 06 '20 at 03:56
  • Yet another way is to notice that from the well-known formula for the sum of integers we have $n^2(n+1)^2/4 = (1 + 2 + \ldots + n)^2$ which is even iff $(1+2+\ldots +n)$ is even. – Winther Oct 06 '20 at 04:00
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    Just show $n=1,2,5,6$ then $\frac {n^2(n+1)^2}4$ is odd. That's all you have to do. – fleablood Oct 06 '20 at 04:00
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    Note: $\frac {n^2(n+1)^2}4$ being even is the same thing as $n^2(n+1)^2$ being divisible by $8$. Only one of $n$ or $n+1$ is even so we have either $8$ dividing $n^2$ or $8$ dividing $(n+1)^2$. If $8=2^3$ divides $M^2$ what does that say about $M$? What feature do the numbers $3,4$ have that $1,2,5,6$ don't? – fleablood Oct 06 '20 at 05:18

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If $\frac {n^2(n+1)^2}4 =2k$ is even. Then $n^2(n+1)^2 =8k$. Now one of $n$ or $(n+1)$ is odd and the other is even. $8$ divides $n^2(n+1)^2$ so either $8$ divides $n^2$ or $8$ divide $(n+1)^2$.

If $8$ divides $n^2$ what does that say about the divisibility of $n$? What does $4$ have that $2$ and $6$ do not?

If $8$ divides $(n+1)^2$ what doest that say about the divisibility of $n+1$? What does $3+1$ have that $1+1$ and $5+1$ do not?

fleablood
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