Difference between revisions of "2010 IMO Problems/Problem 3"
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Lemma 1) <math>g(m) \ne g(m+1)</math> | Lemma 1) <math>g(m) \ne g(m+1)</math> | ||
| − | Assume for contradiction that <math>g(m) = g(m+1)</math> | + | Assume for contradiction that <math>g(m) = g(m+1)</math> |
<math>\left(g(m+1)+m\right)\left(g(m)+m+1\right)</math> has to be a perfect square | <math>\left(g(m+1)+m\right)\left(g(m)+m+1\right)</math> has to be a perfect square | ||
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Lemma 2) <math>|g(m)-g(m+1)| = 1</math> (we have show that it can't be 0) | Lemma 2) <math>|g(m)-g(m+1)| = 1</math> (we have show that it can't be 0) | ||
| − | Assume for contradiction, that <math>|g(m)-g(m+1)| > 1</math>. Then there must exist a prime number <math>p</math> such that <math>g(m)</math> and <math>g(m+1)</math> are in the same residue class modulo <math>p</math>. | + | Assume for contradiction, that <math>|g(m)-g(m+1)| > 1</math>. Then there must exist a prime number <math>p</math> such that <math>g(m)</math> and <math>g(m+1)</math> are in the same residue class modulo <math>p</math>. |
If <math>|g(m)-g(m+1)| = p^aq</math> where <math>q</math> is not divisible by <math>p</math>. | If <math>|g(m)-g(m+1)| = p^aq</math> where <math>q</math> is not divisible by <math>p</math>. | ||
<br /> | <br /> | ||
| − | If <math>a=1</math>. | + | If <math>a=1</math>. |
| − | Consider an <math>n</math> such that <math>g(m)+n =p^3</math> | + | Consider an <math>n</math> such that <math>g(m)+n =p^3</math> |
| − | <math>g(m+1)+n = p^3 \pm pq =p (r)</math> , where <math>r</math> is not divisible by <math>p</math> | + | <math>g(m+1)+n = p^3 \pm pq =p (r)</math> , where <math>r</math> is not divisible by <math>p</math> |
<br /> | <br /> | ||
| − | If <math>a>1</math>. | + | If <math>a>1</math>. |
| − | Consider an <math>n</math> such that <math>g(m)+n =p</math> | + | Consider an <math>n</math> such that <math>g(m)+n =p</math> |
| − | <math>g(m+1)+n = p \pm p^aq =p (r)</math> , where <math>r</math> is not divisible by <math>p</math> | + | <math>g(m+1)+n = p \pm p^aq =p (r)</math> , where <math>r</math> is not divisible by <math>p</math> |
<br /><br /> | <br /><br /> | ||
| − | At least one of <math>g(n)+m</math> , <math>g(n)+m+1</math> is not divisible by <math>p</math>. Hence, | + | At least one of <math>g(n)+m</math> , <math>g(n)+m+1</math> is not divisible by <math>p</math>. Hence, |
| − | At least one of <math>(g(m+1)+n )(g(n)+m +1)</math>, <math>(g(m)+n )(g(n)+m)</math> is divisible by an odd amount of <math>p</math>. | + | At least one of <math>(g(m+1)+n )(g(n)+m +1)</math>, <math>(g(m)+n )(g(n)+m)</math> is divisible by an odd amount of <math>p</math>. |
Hence, that number is not a perfect square. | Hence, that number is not a perfect square. | ||
<br /><br /> | <br /><br /> | ||
| − | Thus, <math>g(x)=x+k</math>, <math>k\in\mathbb{N}</math> | + | Thus, <math>g(x)=x+k</math>, <math>k\in\mathbb{N}</math> |
Revision as of 23:37, 23 October 2010
Problem
Find all functions 
such that 
is a perfect square for all 
Author: Gabriel Carroll, USA
Solution
Suppose such function 
exist then:
Lemma 1) 
Assume for contradiction that
has to be a perfect square
has to be a perfect square
but 
.
A square cannot be between 2 consecutive squares. Contradiction. Thus,
Lemma 2) 
(we have show that it can't be 0)
Assume for contradiction, that. Then there must exist a prime number
such that
and
are in the same residue class modulo
. Then there must exist a prime number 
and 
are in the same residue class modulo If 
where 
is not divisible by 
.
If.
.
Consider ansuch that
such that 
, where
is not divisible by
, where 
is not divisible by
If.
.
Consider an
, where
, where
At least one of,
is not divisible by
, 
is not divisible by At least one of,
is divisible by an odd amount of
, 
is divisible by an odd amount of Hence, that number is not a perfect square.
Thus,,
, 
