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I am trying to work through some of the problems in Stephen Lay's Introduction to Analysis with Proof before my Real Analysis class in the fall term starts, and I was just wondering if I could get some feedback as to whether or not I have completed this proof properly

The question:

Prove that if $n$ is an integer, then $n^2 + n^3$ is an even number

MathMajor
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    Note that $n^2(1+n)=2k+1$ does not imply that either $n^2=2k+1$ or $n+1=2k+1$. Also, $\sqrt{2k+1}$ is sometimes an integer, let $k=4$ for example. And you are not trying to prove that $n$ is even, you are trying to prove that $n^2+n^3$ is even. – André Nicolas Aug 16 '14 at 02:25

7 Answers7

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Your approach for the proof is incorrect. You want to prove an implication $P \to Q$. So if you assume that $Q$ is true, then you have nothing left to prove. You should just begin with an integer $n$ then try to see what can be said about $n^3+n^2$. For example, you can say $$n^3+n^2=n^2(n+1).$$ Now if $n$ is odd then $n+1$ is even and vice versa. So you have a product of odd number and even number and hence even.

Anurag A
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  • I just realized that I misread the question when I made the answer!! – MathMajor Aug 16 '14 at 02:16
  • Unfortunately no. In the line where you say $n^2(n+1)=2k+1$, you cannot conclude that $n^2=2k+1$. – Anurag A Aug 16 '14 at 02:20
  • If you strictly want to use basic definition, then you need to break it into two cases. Let $n$ be even, then $n=2a$ for some $a \in \mathbb{Z}$. Then you can say $n^3+n^2=8a^3+4a^2=4a^2(2a+1)=2[2a^2(2a+1)]$. Now this is of the form $2k$. – Anurag A Aug 16 '14 at 02:22
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Hint

$$n^3+n^2=n(n(n+1))$$

$n(n+1)$ is the product of two consecutive integers.

N. S.
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  • Nice solution! I didn't think of that. – layman Aug 16 '14 at 02:17
  • Question about that, is there any way you can prove that $n^3 + n^2$ is of the form 2$k$? – MathMajor Aug 16 '14 at 02:20
  • @Toxicz The idea is that $n^{3} + n^{2}$ is equal to $n(n(n + 1))$. So we have $nn(n + 1)$. One of those is even because we have both a product of both $n$ and $n + 1$. If you think about it, every other integer is odd, and every other one is even. For example, in order, we have $1,2,3,4,5,6,\cdots$. The $1, 3, 5, \cdots$ are odd and the $2, 4, 6, \cdots$ are even. Every other one is odd. So when you have $n$ and $n + 1$, since they are consecutive and every other integer is even, one of these integers is even. So $nn(n + 1) = 2k$ for some integer $k$. Does that help? – layman Aug 16 '14 at 03:22
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Alternatively, by Fermat's Little Theorem,

$$\begin{align}n^2 + n^3 &= n^2 + n\cdot n^2\\ &\equiv n + n\cdot n \pmod 2\\ &= n + n^2 \pmod 2\\ &\equiv n + n \pmod 2\\ &= 2n \pmod 2\\ &\equiv 0 \pmod 2\end{align}$$

Yiyuan Lee
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  • Or perhaps

    $$\begin{align}n^2+n^3&=n^2(n+1)(\mod 2)\ &=n(n+1)(\mod 2)\ &=0(mod 2)\end{align}$$

     – Hypergeometricx Aug 16 '14 at 05:51
  • Or perhaps $$\begin{align} n^2+n^3&=n^2(1+n)\mod 2\ &=n(n+1)\mod 2\ &=0\mod 2 \end{align}$$

    [couldn't edit the previous comment further... this corrects the typesetting]

     – Hypergeometricx Aug 16 '14 at 06:10
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Here's a shorter proof: Select any $n \in \mathbb{Z}$. There are two cases: $n \equiv 0, 1 \pmod{2}$.

In either case, use modular arithmetic to compute $n^2 + n^3 \pmod{2}$. What can we conclude?

Kaj Hansen
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  • Well, I know what $x$ is congruent to $y$ mod $a$ means, but how would I solve an equation with that sort of "constraint"? – MathMajor Aug 16 '14 at 02:14
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    Well, when we say $x \equiv y \pmod{n}$, then that means that $n|(x-y)$. When we say $n|(x-y)$, that means there is some other integer $z$ such that $nz = x-y$. Therefore, any even number will be equivalent to $0 \pmod{2}$. From here, try to prove the facts that $x \equiv y \pmod{n}$ and $a \equiv b \pmod{n} \implies xa \equiv yb \pmod{n}$, and likewise for addition. The ultimate goal is to show that $n^2 + n^3$ is even, and thus equivalent to $0 \pmod{2}$. – Kaj Hansen Aug 16 '14 at 02:17
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    Thats quite interesting, I will have to read up on this. Thank you :) – MathMajor Aug 16 '14 at 02:21
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This came up as a question that might have the answer to my question. It doesn't, but since I'm here, I might as well:

Suppose $n$ is odd. Then $n^2$ is also odd, as well as $n^3$. But two odd numbers add up to an even number, hence $n^2 + n^3$ is even.

If instead $n$ is even, then $n^2$ and $n^3$ are both even also, and since even numbers add up to even numbers anyway, we're done.

Mr. Brooks
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Integer can be separated into two types one is even number and the other is odd numbers Case 1: The integer is even:

Even numbers can be represented as: $n = 2k$

$n^2 + n^3 = (2k)^2 + (2k)^3 = 4k^2 + 8k^3 = 2(2k^2 + 4k^3) = 2m$

Therefore if n is even $->$ ${n^2 + n^3}$ is even

Case 2: The integer is odd:

Odd numbers can be represented as $n = 2k + 1$

$n^2 + n^3 = (2k + 1)^2 + (2k + 1)^3 = 8k^3 + 16k^2 + 10k + 2 = 2(4k^3 + 8k^2 + 5k + 1) = 2x$

Therefore if $n$ is odd $->$ $n^2 + n^3$ is even

Venkada Narasiman
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My attempt via unique prime factorization of integers:

$n = p_1^{a_1} p_2^{a_2} \dots p_n^{a_n}$ for some unique choice of (positive) prime $p_1, \dots, p_n$ and natural $a_1, \dots, a_n$.

Case 1: Let $n$ be odd. Then each $p_i$, with $1 \leq i \leq n$ is odd (i.e., not 2). Now, $n^2 = (p_1^{a_1} p_2^{a_2} \dots p_n^{a_n})^2 = p_1^{2a_1} p_2^{2a_2} \dots p_n^{2a_n}$, which is still odd. Similar for $n^3$. Thus $n^2 + n^3$ is even, being the sum of two odd numbers.

Case 2: Let $n$ be even. Then we know that one of the $p_i$'s is $2$. We can write our unique factorization of $n$ as $2^{a_1} p_2^{a_2} \dots p_n^{a_n}$. Now it should be obvious that $n^2 = 2^{2a_1} p_2^{2a_2} \dots p_n^{2a_n}$ is even, since it's a multiple of 2, and likewise for $n^3$. Thus we see that $n^2 + n^3$ is the sum of two even integers and is therefore even.

syusim
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