Finding the Intersection of Sets X and Y with Algebraic Methods

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The discussion focuses on finding the intersection of two sets defined by algebraic expressions: X = {x: x=4n+1} and Y = {y: y=m^2+m+1}. An initial attempt to equate the two expressions led to the solutions {1, 13}, but further exploration revealed a larger set of common elements, including {1, 13, 21, 57, 73, 133, 157, 241, 273, ...}. The conversation highlights the importance of considering all possible values for n and m, and a more comprehensive algebraic approach was suggested, leading to the conclusion that the intersection can be expressed as {x: x=m^2 + m + 1; m/4∈N or (m+1)/4∈N}. This method emphasizes the need to analyze both sets thoroughly to find all common elements.
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Homework Statement


X = {x: x=4n+1, n\inN}
Y = {y: y=m^{2}+m+1, m\inN}

Find: Y\capX

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 into 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?

Thanks in advance,
Yorick.
 
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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?
 
Yes, Thankyou!

So basically the solution is something like {x: x=m^2 + m + 1 ; m/4\inN or (m+1)/4\inN}
 
yorick said:
Yes, Thankyou!

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

yep... that's right!
 
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