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Topic: Root + Root = Integer (Read 771 times) |
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ThudnBlunder
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Root + Root = Integer
« on: Mar 18th, 2009, 9:33pm » |
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If p and q are prime numbers such that  (p2 + 7pq + q2) + (p2 + 14pq + q2) is an integer, prove that p and q must be equal.
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| « Last Edit: Mar 19th, 2009, 4:20am by ThudnBlunder » |
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Aryabhatta
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Re: Root + Root = Integer
« Reply #1 on: Mar 18th, 2009, 10:10pm » |
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on Mar 18th, 2009, 9:33pm, ThudanBlunder wrote:| If p and q are prime numbers such that (p2 + 7pq + p2) + (p2 + 14pq + p2) is an integer, prove that p and q must be equal. |
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Did you mean (p2 + 7pq + q2) + (p2 + 14pq + q2) ?
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| « Last Edit: Mar 18th, 2009, 10:11pm by Aryabhatta » |
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ThudnBlunder
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Re: Root + Root = Integer
« Reply #2 on: Mar 19th, 2009, 4:19am » |
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on Mar 18th, 2009, 10:10pm, Aryabhatta wrote:
Did you mean (p2 + 7pq + q2) + (p2 + 14pq + q2) ?
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Yes, thank you. (Corrected)
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FiBsTeR
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Re: Root + Root = Integer
« Reply #3 on: Mar 29th, 2009, 6:50pm » |
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I don't want this to get buried so here's what I have so far:
LEMMA: The radicands are perfect squares.
Proof: If we let the roots be a,b, we have a2,b2,a+b are rational, so a-b=(a2-b2)/(a+b) is rational also. Therefore a,b are rational. But a2,b2 are integers, so a,b must be integers also. [end lemma]
Now let:
m2 = p2 + 14pq + q2
n2 = p2 + 7pq + q2
--> m2-n2=(m+n)(m-n)=7pq
The p=q=2 case works, and in all other cases p,q are odd primes >=3. Then since m+n>m-n, we have only the cases:
m-n=1, m+n=7pq
m-n=p, m+n=7q
m-n=q, m+n=7p
m-n=7, m+n=pq
I believe the middle two cases lead to p=q when p,q are related in a quadratic and the discriminant checked for squareness, but I don't know what to do with the other two.
----
I really should work on my number theory for the USAMO next month. 
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Hippo
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Re: Root + Root = Integer
« Reply #4 on: Mar 30th, 2009, 3:06am » |
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Why is m-n=7p,m+n=q excluded?
I get
Suppose d=q-p>0. 16p(p+d)=m2-d2=(m+d)(m-d). Possibilities are:
4p=m-d,4(p+d)=m+d giving d=0 contradiction
2p=m-d,8(p+d)=m+d giving d=-p contradiction
2(p+d)=m-d,8p=m+d giving d=3/2p contradiction
as m+d has same parity as m-d and p,(p+d) are odd primes there are no other possibilities.
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| « Last Edit: Mar 30th, 2009, 4:27am by Hippo » |
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Vondell
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Re: Root + Root = Integer
« Reply #5 on: Mar 30th, 2009, 7:48am » |
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"1" works in the formula just as well. However, other than inserting prime numbers at random, I have absolutely no idea how to "prove" the rule/theorem.
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Peterman
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Re: Root + Root = Integer
« Reply #6 on: Apr 1st, 2009, 2:14am » |
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As FiBsTeR pointed out, the square-rooted expressions must be perfect squares. So:
p2+7pq+q2 is a square, a bit larger than (p+q)2: call it (p+q+n)2
So 5pq = n(2p+2q+n), hence n|5pq, thus there are eight possibilities: n=1,5,p,q,5p,5q,pq or 5pq.
Each of these leads to either no solution or to p=q.
For example, n=5 leads to (p-2)(q-2)=9, so either p=q=5 or else p and q are 3 and 11.
But in the (3,11) case, p2+14pq+q2 = 592 is not a square.
Curious that there seems to be no single reason why p=q, it just works out that way in each case.
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