Simple Number Theory Task

Question

I am doing exercises with middle grade students and looking at their capacities to create arguments for simple number theory conjectures. I want three tasks, and I have two so far: 1. The sum of three consecutive integers is divisible by three. 2. The sum of two odd numbers is an even number. I need one more task that is similar in difficulty, but not too similar that the same warrants would arise. Any ideas?

Answer 1

Maybe discuss the twin primes conjecture briefly and then ask them to investigate how many triplet primes of the form $p,p+2,p+4$ there are. They should be able to prove (using the kinds of arguments you'd use in tasks 1. and 2.) that there is only one such triplet. I think this both shows them that number theory is pretty deep and that conjecture is part of the natural cycle of doing mathematics.

Answer 2

A nice third problem arises by continuing from the second problem to show the power of parity arithmetic for solving problems in integer arithmetic. For example, we can use it to show that large classes of polynomials have no integer roots. Let's consider a simple example.

$f(x) = x(x\!+\!1)+2n\!+\!1\,$ has no integer roots since $\,x(x\!+\!1)\,$ is even, so adding $2n\!+\!1$ yields an odd so $\ne 0;\,$ equivalently $\,f(0)\equiv 1\equiv f(1),\,$ i.e. $f\,$ has no roots mod $\,2,\,$ so no integer roots.

Remark $ $ This idea generalizes widely as follows

Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd.

Proof $\ $ If so then $\ \color{#0a0}{f(0)} \equiv 1\equiv \color{#c00}{f(1)}\,\pmod 2,\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots. $\ $ QED

In the same way, we can often reduce problems to "smaller", simpler problems in modular images. Because (ring) homs preserve the ambient algebraic structure, as above, we can often deduce information about the original problems from information gleaned in the simpler modular images. Such problem solving by modular reduction is an algebraic way of "dividing and conquering".

As above, one can often introduce the germ of such ideas in very early grades through simple instances such as parity arithmetic. When teaching at such early levels I strive to present examples that will provide good intuition for the generalizations that students will encounter in later studies. The Parity Root Test is a model example of such. You can find much further discussion of such in my many MSE posts on the Parity Root Test.

Answer 3

The NRICH website has several posters that I like to pick interesting problems like this from. Right now, these three are my favorite:
1. Big Powers
2. Make 37
3. Mystic Rose
They have more poster problems for secondary and primary ages. The problems are generally simple to state and allow students to play around with the math a bit before making conjectures and proving them.

Answer 4

I like your question a lot :) and got some new ideas from other answers. So I like to contribute some of mine in the hope they provide useful to others as well. I usually distinguish if problems are relating to place value system or not.

1. Palindromic numbers ($ANNA$-numbers). Using two different digits, you can build two different $ANNA$-numbers. For example using $7$ and $3$, you can build $7447$ and $4774$. What can you say (and prove) about their difference?

2. Mirrors. Sometimes, the product of a two digit numbers and their mirrors are equal. Like $23\times 64= 32\times 46$. Or $34\times 86 = 43\times 68.$

Can you figure (and prove) a special property of numbers where this is true?

3. Sum of consecutive numbers. Actually, your first problem generalizes a lot and allows to equal or more challenging problems, like : What numbers can not be written as a sum of consecutive numbers.

Remark-1: Actually, this problem is also on nrich.

4. Gaps in Prime numbers. For each pos. integer $n$ there is a sequence of $n$ numbers, none of which is prime.

Remark-1: this is nice to show that the general solution (i.e. following a rule like take $(n+1)!$, $(n+1)!+1$, $(n+1)!+2$, $\ldots$, $(n+1)!+(n+1)$ does not always produce the easiest example possible, as you can see e.g. setting $n=8$.

Remark-1: If your students know Euclids proof that there are infinitely many primes, you can also do the following problem:

Prove that for $n\geq 3$, there is at least one prime number between $n$ and $n!$.

Answer 5

The product of an even number and an odd number is an even number.

or

The product of two odd numbers is again an odd number.

Answer 6

The divisible by 3 thing (digits addition)?