# Finding unions of sets

#### yorick

1. The problem statement, all variables and given/known data
X = {x: x=4n+1, n$$\in$$$$N$$}
Y = {y: y=m$$^{2}$$+m+1, m$$\in$$$$N$$}

Find: Y$$\cap$$X

3. The attempt at a solution

First I tried to set x=y
$$\Rightarrow$$ n$$^{2}$$+n+1=4n+1
$$\Rightarrow$$ n$$^{2}$$-3n=0
$$\Rightarrow$$ n(n-3)=0
$$\Rightarrow$$ n=0,3

Subbing this in to either equation yields {1,13}.
I thought I was doing okay.
However I was comparing assignments with a classmate and she had this massive set that just keeps on going: {1, 13, 21, 57, 73, 133, 157, 241, 273, ...}
After I thought about it, this made sense, as my solution only found the elements that are in both sets when n is the same for both equations, but set equality relies only on its elements, not at what stage they were reached.
She'd done this the long way round, scripting a little program to calculate all elements for both sets then compare the elements and find the common ones.
What I'd like to know is, how can I go about solving this algebraically?

Yorick.

#### praharmitra

write the equation as it is.

m^2 + m + 1 = 4n + 1
--> m^2 + m = 4n
--> m(m+1) = 4n

now rhs is a multiple of 4... and therefore, so must the lhs. But if m is even, m + 1 is odd...
do you see where i am going with this? can you proceed?

#### yorick

Yes, Thankyou!

So basically the solution is something like {x: x=m^2 + m + 1 ; m/4$$\in$$$$N$$ or (m+1)/4$$\in$$$$N$$}

#### praharmitra

Yes, Thankyou!

So basically the solution is something like {x: x=m^2 + m + 1 ; m/4$$\in$$$$N$$ or (m+1)/4$$\in$$$$N$$}
yep... thats right!

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