Difference between revisions of "2007 Indonesia MO Problems/Problem 8"
Rockmanex3 (talk | contribs) (Solution to Problem 8 (credit to crazyfehmy) -- perfect square shenans) |
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Let <math> m</math> and <math> n</math> be two positive integers. If there are infinitely many integers <math> k</math> such that <math> k^2+2kn+m^2</math> is a perfect square, prove that <math> m=n</math>. | Let <math> m</math> and <math> n</math> be two positive integers. If there are infinitely many integers <math> k</math> such that <math> k^2+2kn+m^2</math> is a perfect square, prove that <math> m=n</math>. | ||
− | ==Solution (credit to crazyfehmy)== | + | ==Solution 1 (credit to crazyfehmy)== |
Note that we can [[Completing the square|complete the square]] to get <math>k^2 + 2kn + n^2 - n^2 + m^2</math>, which equals <math>(k+n)^2 + m^2 - n^2</math>. | Note that we can [[Completing the square|complete the square]] to get <math>k^2 + 2kn + n^2 - n^2 + m^2</math>, which equals <math>(k+n)^2 + m^2 - n^2</math>. | ||
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Now we need to prove that if <math>m = n</math>, there are an infinite number of integers <math>k</math> that satisfy the original conditions. By the Substitution Property, we find that <math>k^2 + 2kn + m^2 = k^2 + 2kn + n^2</math>. The expression can be factored into <math>(k+n)^2</math>. Since the expression is a perfect square, for all integer values of <math>n, k</math>, there are an infinite number of integers <math>k</math> that satisfies the original conditions when <math>m = n</math>. | Now we need to prove that if <math>m = n</math>, there are an infinite number of integers <math>k</math> that satisfy the original conditions. By the Substitution Property, we find that <math>k^2 + 2kn + m^2 = k^2 + 2kn + n^2</math>. The expression can be factored into <math>(k+n)^2</math>. Since the expression is a perfect square, for all integer values of <math>n, k</math>, there are an infinite number of integers <math>k</math> that satisfies the original conditions when <math>m = n</math>. | ||
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+ | ==Solution 2 (credit to dskull16)== | ||
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+ | We begin by completing the square to get <math>k^2 + 2kn + n^2 - n^2 + m^2</math>, which equals <math>(k+n)^2 + m^2 - n^2</math>. | ||
+ | |||
+ | <br> | ||
+ | Then we have that <math>(k+n)^2 + m^2 - n^2 = a^2</math> for some natural number a. | ||
+ | This then gives us <math>m^2 - n^2 = a^2 - (k+n)^2</math> which we can write like | ||
+ | <math>(m+n)(m-n) = (a+k+n)(a-k-n)</math> by difference of two squares. | ||
+ | |||
+ | <br> | ||
+ | Now we remark that the left hand side is a constant since we prematurely chose <math>m</math> and <math>n</math>. Acknowledging the fact that this equation is comprised entirely of integers, we see that <math>(a+k-n)</math> and <math>(a-k-n)</math> need both be factors of the left hand side of which there are finitely many. This means that there are finitely many solutions for <math>a</math>. | ||
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+ | <br> | ||
+ | If however the left hand side were 0, implying that either <math>m = n</math> or <math>m = -n</math>, we would be able to find infinitely many integers such that <math>a+k = n</math>. Since <math>m</math> and <math>n</math> are positive integers, this means that <math>m=n</math> as required. | ||
==See Also== | ==See Also== |
Latest revision as of 17:08, 3 February 2024
Contents
[hide]Problem
Let and be two positive integers. If there are infinitely many integers such that is a perfect square, prove that .
Solution 1 (credit to crazyfehmy)
Note that we can complete the square to get , which equals .
Assume that . Since are positive, we know that . In order to prove that is not a perfect square, we can show that there are values of where .
Since , we know that . In the case where , we can expand and simplify to get
All steps are reversible, so there are values of where , so there are no values of where that results in infinite number of integers that satisfy the original conditions.
Now assume that . Since are positive, we know that . In order to prove that is not a perfect square, we can show that there are values of where .
Since , we know that . In the case where , we can expand and simplify to get
All steps are reversible, so there are values of where , so there are no values of where that results in infinite number of integers that satisfy the original conditions.
Now we need to prove that if , there are an infinite number of integers that satisfy the original conditions. By the Substitution Property, we find that . The expression can be factored into . Since the expression is a perfect square, for all integer values of , there are an infinite number of integers that satisfies the original conditions when .
Solution 2 (credit to dskull16)
We begin by completing the square to get , which equals .
Then we have that for some natural number a.
This then gives us which we can write like
by difference of two squares.
Now we remark that the left hand side is a constant since we prematurely chose and . Acknowledging the fact that this equation is comprised entirely of integers, we see that and need both be factors of the left hand side of which there are finitely many. This means that there are finitely many solutions for .
If however the left hand side were 0, implying that either or , we would be able to find infinitely many integers such that . Since and are positive integers, this means that as required.
See Also
2007 Indonesia MO (Problems) | ||
Preceded by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Last Problem |
All Indonesia MO Problems and Solutions |