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x−1x5−1=x−1(x−1)(x4+x3+x2+x+1)=5100+575+550+525+1=y but 3597751∣y∴notprime You also might be able to use the pseudo prime test i.e. ay−1=1mod(y)forgcd(a,n)=1
@Shashank Rammoorthy
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Yeah. I thought that difference of two squares may be the way to go, so I got (Ax2+Bx+1)2 and, through trial and error, found (fortunately) appropriate values for A and B so that it worked. It could have easily not worked either.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
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x−1x5−1=x−1(x−1)(x4+x3+x2+x+1)=5100+575+550+525+1=y but 3597751∣y∴ not prime You also might be able to use the pseudo prime test i.e. ay−1=1mod(y) for gcd(a,n)=1
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FYI For exponents with multiple digits, use { } for all of them to display properly, EG 5^{100}
How did you guess that 3597751∣y? Or did you use a computer?
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yea I used wolfram alpha
Haha. I guess this is a valid proof :) I proved it using difference of two squares.
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Let x=525
Then 525−15125−1=x−1x5−1=x4+x3+x2+x+1
You can notice that x4+x3+x2+x+1=(x2+3x+1)2−(5x3+10x2+5x)
=(x2+3x+1)2−5x(x2+2x+1)
=(x2+3x+1)2−526(x+1)2
And then, using difference of two squares, you get:
((x2+3x+1)−513(x+1))((x2+3x+1)+513(x+1))
So, the above expression has these two factors, hence it is not prime.
Q.E.D
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x4+x3+x2+x+1=(x2+3x+1)2−(5x3+10x2+5x), though? Was it purely intuition?
Cool proof. How'd you getLog in to reply
(Ax2+Bx+1)2 and, through trial and error, found (fortunately) appropriate values for A and B so that it worked. It could have easily not worked either.
Yeah. I thought that difference of two squares may be the way to go, so I gotLog in to reply
Nice. Here is a similar question :)