Classic NT Problem
For how many positive integers n ≤ 2 0 1 4 is the expression
n 7 + 7
a perfect square?
Furthermore, as this is an olympiad-type problem, can you prove these are the only solutions?
The answer is 0.
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3 solutions
The correct answer is 29. The above is a perfect square for all squares of n >= 16
Correct answer is 29
1 Number is =256 2 Number is =289 3 Number is =324 4 Number is =361 5 Number is =400 6 Number is =441 7 Number is =484 8 Number is =529 9 Number is =576 10 Number is =625 11 Number is =676 12 Number is =729 13 Number is =784 14 Number is =841 15 Number is =900 16 Number is =961 17 Number is =1024 18 Number is =1089 19 Number is =1156 20 Number is =1225 21 Number is =1296 22 Number is =1369 23 Number is =1444 24 Number is =1521 25 Number is =1600 26 Number is =1681 27 Number is =1764 28 Number is =1849 29 Number is =1936 Thanks
Write the equation in the form: n^7 + 128 = m^2 + 121, factorize the LHS and use modulo 4, with the theorem for p = 4k+3, p/a^2 + b^2 => p/a, p/b :)
Interesting ideas. Can you elaborate on the procedure and provide more details?
Try n=36. I believe that 36^7 + 7 is a perfect square. Plus many other values for n.
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36^7+7 has 3 at unit's place. It can't be a perfect square.
Can someone please provide a solution?I did some mod bashing but it didn't work quite well.Jon Haussmann,Pi Han Goh,Sharky?Finn?
I read a solution in yahoo but I'm stuck at Z[stuffs] that I don't know what it is T__T.
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Z4 means "the integers modulo 4"
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Is it like n 2 ≡ 0 , 1 ( m o d 4 ) ? And if yes, why did we use Z4 or stuffs and not just using congruence?
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@Samuraiwarm Tsunayoshi – It's the same, just more formal.
I'm interested in that link. Did you look that up to solve the problem?
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I guessed that it has to be 0 because n 3 + 7 can't be perfect squares. Yep my guess is really pointless =="
How do you tag people?
Did you just guess the answer?
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Kind of :D.I thought my solution was correct, then found a mistake and I was wondering how to solve it.I guess it's either a factoring trick or mod bashing.
It's not too hard but I'll mention those people: @Jon Haussmann @Pi Han Goh @Sharky Kesa .
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I respectfully defer to the proposer to post his solution.
trial and error
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For 2014 numbers?!Geez, you must have a lot of spare time :D
As the expression has to be a perfect square,therefore we get the following 2 equations-
n 7 is either 0 or 1 congruent modulo 4
or 4| n 7 +7 and again 4| n 7 +6
or 4| n 7 +7-( n 7 +6)
or 4|1.
this is not possible and so there are no perfect squares of the type n 7 +7.