Prove that x^2 + x is even for every integer x

Prove that x^2 + x is even for every integer x

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[F.E.Terman]

x(x+1) is the easiest. But, for fun, we can also use induction:
Suppose for some x, _x²+x_ is even. Then, (x+1)² + (x+1) = x²+2x+1 + x+1 = _x²+x_ + 2(x+1), which is even plus even, so, even.
Likewise, (x-1)² + (x-1) =x²-2x+1 + (x-1) = _x²+x_ -2x, which is even minus even, so, even.
So if x²+x is even for _some_ x, then for _all_ x. Now prove that it is for some x: just for the fun of it, choose x=7. 49+7 = 56 is even, therefore, all x²+x is even. 👌

[rampakeshbharat]

a lot less work if problem is written as x( x + 1) at the start...

[barryzeeberg]

much simpler:
x^2 + x = x(x + 1)
either x is even and x+1 is odd, or vice versa.
in either case, even * odd = even

[ILoveMaths]

x is either odd or even.
If x is odd, x^2 is odd, and odd + odd (x^2 + x) = even.
If x is even, x^2 is even, and even + even (x^2 + x) = even.
QED

[seegeeaye]

Induction can prove it easily, or use this way: x^2+x = x(x+1), we are proving the product of two consecutive integers is even, i.e. 2m(2m +1) is even

[Penrose]

I love substitution based proofs like these for number theory problems. They feel very "organic" for some reason. Keep em' coming Professor, thank you! This question in particularly is lovely because of the use of the pronic identity x(x+1). Now that we know that this identity produces exclusively even integral results for all integral x, then we may set the RHS equal to 2k without any restrictions on our range. Solving for k now yields the equation for triangular numbers, T = 2 | x(x+1) = (x^2+x)/2. Perhaps there is not a set which I love more than the triangle numbers.

[Fircasice]

For me, one key takeaway here is that even if you don't see the more elegant solution to a problem, simply by following mathematical methodology, you can still prove things and find solutions.

[lolzhunter]

first we assume (x^2+x) is even
(x^2+x)+2x-1 = odd number since 2x-1 is the definition of an odd number and also even+odd is always odd
left with x^2+3x-1 is odd
therefore x^2+3x must be even
in order for this to be true, both x^2 and 3x must be both odd or even for every integer x
going through 3x, the last digit repeats after x=10
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
we can simplify this into 3, 6, 9, 2, 5, 8, 1, 4, 7, 0 since we only really care about the last digit
now we look at x^2, x^2 also repeats its last digit after x=10
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
simplified that is 1, 4, 9, 6, 5, 6, 9, 4, 1, 0
matching them up we get
3, 6, 9, 2, 5, 8, 1, 4, 7, 0
1, 4, 9, 6, 5, 6, 9, 4, 1, 0
doing a quick check you can see that every pair of numbers (3, 1) (6, 4) (9, 9) etc are both odd or even
this means that when added together x^2 and 3x will always be even for any x when x is an integer
so backtracking a bit if x^2+3x is even, then x^2+3x-1 has to be odd by definition
which verifies the earlier statement we made that (x^2+x)+2x-1 = odd number, and as we know from before adding an odd number and an even number together will get you an odd number, and we know 2x-1 is odd by definition, this then means that (x^2+x) has to be even by proxy, therefore x^2+x is even

(tell me what you think of this proof in the replies, i came up with it after seeing the title before i watched the video so its a bit all over the place, also if i messed up anywhere then please tell me lol)

EDIT: on further inspection i kind of went a bit backwards, i could have applied the same number checking to just x^2 and x and got a quicker result but hey whatever

[SSJ]

x^2 + x
= x(x+1)
either x or x+1 is even
anything multiplied by an even number is itself even
therefore x^2 + x is even

[SmartLizardTutorials]

Ofcourse it will always work because (even + even = even)
and (odd + odd = even)
Also (even^2 = even)
and (odd^2 = odd)

Therefore, there can never be a moment where even will be added to an odd or vice versa.

[fraternitas]

This is exactly the kind of content I want to see, gratias tibi ago!

[FlaminTubbyToast]

An interesting way to prove this would be to prove that x^2 maintains parity and then because x almost by definition maintains parity then in both cases x^2 + x must be even

[gackerman]

love videos like this - accessible tastes of things

[writerightmathnation]

I wish you'd mention the prerequisites for your videos. Your solution is great for k12 students, but there are more insightful arguments available to reasonably educated sophomores. I love your book reviews, though, and your self study suggestion videos.

[DevsTapePlayer]

proofs are some of the most elegant things to watch. I recently began going through Euclid's Elements for that very reason even though it is not taught anymore. That very first proof in book 1 is transformative.

[DavesMathVideos]

Made a similar video about x^4-x^2. Application of the pigeonhole principle.

[koharaisevo]

Reading the title I though you're going to prove x^2+x is an even fuction and I were like how is that possible. 😅

[Bob-zgzf]

Would you please talk about Kalman Filter? It involves lots of advanced math and it is applied in many engineering fields.
Any introductions would be highly appreciated.

[saarlevy]

When I saw the thumbnail my mind immediately went to functions, as in f(x) = x^2 + x is even. And I was like "What the fuck is this guy talking about"