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

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(Original post by

Hard questions on what?

**J.F.N**)Hard questions on what?

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#4

Show that the cube roots of three distinct prime numbers cannot be terms in an arithmetic progression

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#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)

- 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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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.

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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#7

(Original post by

If (a^2+b^2)/(1+ab) is an integer, show that it must be a perfect square.

**trigtrig**)If (a^2+b^2)/(1+ab) is an integer, show that it must be a perfect square.

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#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

(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

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#9

(Original post by

Show that the cube roots of three distinct prime numbers cannot be terms in an arithmetic progression

**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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#10

(Original post by

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

**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

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#11

Here's one:

What is the greatest integer that divides (p^4 - 1), where p is prime and greater than 5.

What is the greatest integer that divides (p^4 - 1), where p is prime and greater than 5.

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#12

(Original post by

Here's one:

What is the greatest integer that divides (p^4 - 1), where p is prime and greater than 5.

**dvs**)Here's one:

What is the greatest integer that divides (p^4 - 1), where p is prime and greater than 5.

Did you have something else in mind Or did you mean for all such p?

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#14

(Original post by

a² - qba + (b² - q) = 0

Quadratic in a and call the other root a'.

**SsEe**)a² - qba + (b² - q) = 0

Quadratic in a and call the other root a'.

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#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.

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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