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Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 9"

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==Solution==
 
==Solution==
 
<math>a=6;b=19,c=30</math> (also <math>2-34-47</math>)
 
<math>a=6;b=19,c=30</math> (also <math>2-34-47</math>)
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Let <math>a+b = x^2</math>, <math>a + c = y^2</math>, <math>b + c = z^2</math>. We can easily find that <math>a = \dfrac{x^2+y^2-z^2}{2}</math>, <math>b = \dfrac{x^2+z^2-y^2}{2}</math>, <math>c = \dfrac{y^2+z^2-x^2}{2}</math>. Taking mod 2, we find that <math>(x, y, z)</math> must be either <math>(0, 0, 0)</math>, <math>(0, 1, 1)</math>, <math>(1, 0, 1)</math>, <math>(1, 1, 0)</math>. For the first case, we check <math>(2n, 2n + 2, 2n + 4)</math> until <math>(a, b, c)</math> is positive. We get <math>n = 4</math>, and <math>(a, b, c) = (10, 54, 90)</math>. For the second case, we check <math>(2n, 2n+1, 2n+3)</math>, getting <math>n = 3</math>, and <math>(a, b, c) = (2, 34, 47)</math>. For the third case, we check <math>(2n + 1, 2n + 2, 2n + 3)</math>, getting <math>n = 2</math> and <math>(a, b, c) = (6, 19, 30)</math>. For the last case, we check <math>(2n + 1, 2n + 3, 2n + 4)</math>, getting <math>n = 2</math>, and <math>(a, b, c) = (5, 20, 44)</math>. Clearly, our triple with the minimum <math>c</math> value is <math>(6, 19, 30)</math>.
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~Puck_0
  
 
==See Also==
 
==See Also==

Latest revision as of 20:30, 29 April 2024

Problem

Determine three positive integers $a,b$ and $c$ that simultaneously satisfy the following three conditions:

(i) $a<b<c$

(ii) Each of $a+b,a+c$ and $b+c$ is the square of an integer, and

(iii) $c$ is as small as is possible.

Solution

$a=6;b=19,c=30$ (also $2-34-47$)

Let $a+b = x^2$, $a + c = y^2$, $b + c = z^2$. We can easily find that $a = \dfrac{x^2+y^2-z^2}{2}$, $b = \dfrac{x^2+z^2-y^2}{2}$, $c = \dfrac{y^2+z^2-x^2}{2}$. Taking mod 2, we find that $(x, y, z)$ must be either $(0, 0, 0)$, $(0, 1, 1)$, $(1, 0, 1)$, $(1, 1, 0)$. For the first case, we check $(2n, 2n + 2, 2n + 4)$ until $(a, b, c)$ is positive. We get $n = 4$, and $(a, b, c) = (10, 54, 90)$. For the second case, we check $(2n, 2n+1, 2n+3)$, getting $n = 3$, and $(a, b, c) = (2, 34, 47)$. For the third case, we check $(2n + 1, 2n + 2, 2n + 3)$, getting $n = 2$ and $(a, b, c) = (6, 19, 30)$. For the last case, we check $(2n + 1, 2n + 3, 2n + 4)$, getting $n = 2$, and $(a, b, c) = (5, 20, 44)$. Clearly, our triple with the minimum $c$ value is $(6, 19, 30)$. ~Puck_0

See Also

2006 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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