# Can someone post some hard questions on here please?Watch

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#1
Thank you
0
14 years ago
#2
Hard questions on what?
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#3
(Original post by J.F.N)
Hard questions on what?
i dunno, number theory. I've tried putting problems up for discussion before, but hardly anyone replies.
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14 years ago
#4
Show that the cube roots of three distinct prime numbers cannot be terms in an arithmetic progression
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14 years ago
#5
Ok, here are some elementary number theory problems:

- Let GCD (a,b)=1, a>0, b>0. Prove that every integer greater than ab-a-b is representable as ax+by (x>=0,y>=0). Moreover, prove that ab-a-b is not representable as such.

- Let log(1025/1024)=a, log((1024^2)/(1023.1025))=b, log((81^2)/(80.82))=c, log((125^2)/(124.126))=d, log((99^2)/(98.100))=e. Show that

196log2=59+5a+8b-3c-8d+4e

Further, express log3 and log41 in terms of a,b,c,d,e. Determine log2 to ten decimal places and discuss the practical application of the method. (All logs are base 10. You are given that log_e(10)=2.3025850930)
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#6
Thanks. While im doing those, i've got some for you:

Show that (a+b , (a^p + b^p)/(a+b)) = 1 or p where a and b are pos. integers and (a,b) = 1, and p is an odd prime

If (a^2+b^2)/(1+ab) is an integer, show that it must be a perfect square.
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14 years ago
#7
(Original post by trigtrig)
If (a^2+b^2)/(1+ab) is an integer, show that it must be a perfect square.
a,b positive integers?
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14 years ago
#8
Yeah. a and b are positive integers. The solution is something to the tune of:

(a²+b²)/(1+ab) = q

Assume we have a minimum solution (a, b) where b=<a
a² - qba + (b² - q) = 0
Quadratic in a and call the other root a'.
By comparing coefficients:
a + a' = qb
aa' = b² - q

Now a' must be an integer as qb = a + a' is an integer.

From the second equation we have:
a' = b²/a - q/a < a
Assume a'>0
As b=<a then b²/a =< a and so b²/a - q/a < a
But that means (a', b) is a new minimum solution. So a' cannot be positive.

(a+1)(a'+1) = aa' + a + a' + 1 = b² + q(b-1) + 1 > 0 (as b is a positive integer)
So a'>-1. But a' is an integer so a' = 0

aa' = b² - q
0 = b² - q
q = b²
0
14 years ago
#9
(Original post by mik1w)
Show that the cube roots of three distinct prime numbers cannot be terms in an arithmetic progression

a, b and c are the 3 primes. Assume that cbrt[a], cbrt[b] and cbrt[c] are part of an arithmetic series with common difference d. Then:
cbrt[c] = md + cbrt[b] => (cbrt[c] - cbrt[b])/m = d, where m is some integer.
cbrt[c] = nd + cbrt[a] => (cbrt[c] - cbrt[a])/n = d, where n is some integer.
So:
(cbrt[c] - cbrt[b])/m = (cbrt[c] - cbrt[a])/n => (cbrt[c] - cbrt[b])/(cbrt[c] - cbrt[a]) = m/n
So, from our assumption, (cbrt[c] - cbrt[b])/(cbrt[c] - cbrt[a]) is rational. A little manipulation:
(cbrt[c] - cbrt[b])/(cbrt[c] - cbrt[a]) = m/n
ncbrt[c] - ncbrt[b] = mcbrt[c] - mcbrt[a]
cbrt[c](n-m) + mcbrt[a] = ncbrt[b]
cbrt[c](n-m)/n + (m/n)cbrt[a] = cbrt[b]
Pcbrt[c] + Qcbrt[a] = cbrt[b], where P and Q are rational.
Cube to get:
b = P³c + Q³a + 3PQ(Pcbrt[c²a] + Qcbrt[ca²])
Now Pcbrt[c²a] + Qcbrt[ca²] = cbrt[ca].cbrt[b] = cbrt[abc], so:
b = P³c + Q³a + 3PQcbrt[abc]
From this we can see that cbrt[abc] is rational, but this is impossible since a, b and c are distinct primes. Contradiction.
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14 years ago
#10
(Original post by trigtrig)
Show that (a+b , (a^p + b^p)/(a+b)) = 1 or p where a and b are pos. integers and (a,b) = 1, and p is an odd prime
What's (a+b , (a^p + b^p)/(a+b))?
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14 years ago
#11
Here's one:
What is the greatest integer that divides (p^4 - 1), where p is prime and greater than 5.
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14 years ago
#12
(Original post by dvs)
Here's one:
What is the greatest integer that divides (p^4 - 1), where p is prime and greater than 5.
The way you've set the question the answer is p^4-1.

Did you have something else in mind Or did you mean for all such p?
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14 years ago
#13
Ah, sorry. I meant for all prime p greater than 5.
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14 years ago
#14
(Original post by SsEe)
a² - qba + (b² - q) = 0
Quadratic in a and call the other root a'.
Are you saying a is a root?
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14 years ago
#15
Yes. If (a,b) is a solution (so a is a constant, not a variable) then plugging a into the quadratic:
x² - qbx + (b² - q)
gives 0 so a is a root.
Sorry. I should have given the solution (a,b) different letters.
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