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kkk579

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no matter how times i read the solution and attempt it i cannot understand why they let k equal 4p, 4p+1, etc. it says n is divided by 4 so why k?Screenshot 2024-10-16 221012.pngScreenshot 2024-10-16 221046.png
 

kkk579

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Also for this q could u apply the concept factor R/S and say that x = plus minus b / plus minis a, but for simplicity we only look at b/a. Then we can say that b /a is essentially (k x p)/(k x q) where k is js a scaling factor. Therefore p can divide b and q can divide a.Screenshot 2024-10-16 230151.pngScreenshot 2024-10-16 230201.png
 
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no matter how times i read the solution and attempt it i cannot understand why they let k equal 4p, 4p+1, etc. it says n is divided by 4 so why k?View attachment 44857View attachment 44858
since k is an integer then it can be written in the form 4m+r, where r is the remainder. we choose k instead of n as n is specifically a square integer (k^2) and if u do substitute it as 4m+r u probably wont be able to easily find the contradiction as u cant really do any working out if u blindly substitute n.
Also for this q could u apply the concept factor R/S and say that x = plus minus b / plus minis a, but for simplicity we only look at b/a. Then we can say that b /a is essentially (k x p)/(k x q) where k is js a scaling factor. Therefore p can divide b and q can divide a.View attachment 44863View attachment 44864
x is a root of the equation and ur assuming that b/a is also a root which is not true.
 

kkk579

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since k is an integer then it can be written in the form 4m+r, where r is the remainder. we choose k instead of n as n is specifically a square integer (k^2) and if u do substitute it as 4m+r u probably wont be able to easily find the contradiction as u cant really do any working out if u blindly substitute n.

x is a root of the equation and ur assuming that b/a is also a root which is not true.
for the 2nd q yeah that does make sense, but for the 1st one the q says that n is getting divided by 4 so i still dont understand why we let n = 4p + r
 

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is this from fort st trial
i did this yesterday lol
 

epicmaths

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Uhm in Q1, letting k be even and k be odd ie k=2p and k=2p+1 works just fine. Not sure the fuss with 4...but this comes from this number theory concept called modulo arithmetic that claims that every integer k can be written as either 4p, 4p+1, 4p+2, 4p+3, because when you reach 4p+4 you looped back to 4(p+1) which is of the first type.
 

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