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Difference between revisions of "1996 IMO Problems/Problem 4"
 
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==Solution==
 
==Solution==
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{{solution}}
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Let us first set up two equations given the information in the problem:
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Let <math>15a + 16b = m^2</math>
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Let <math>16a - 15b = n^2</math>
 +
 
 +
In general, when we are given alternating sums and differences like this, it is a good idea to square both equations and add them up to cancel out any cross terms. Doing this results in the equation <math>m^4 + n^4 = 481(a^2 + b^2)</math>.
 +
 
 +
Note that <math>481</math> factorizes into <math>13 \cdot 37</math>. Thus, <math>m^4 + n^4 \equiv 0\mod 13</math>, and <math>m^4 + n^4 \equiv 0\mod 37</math>. Now, we make a table for all modular congruences possible for m^4 for all nonnegative integers with a unique modulo 13 and 37. Note that due to the modular symmetry of even powers (<math>(-a)^2 = a^2</math>), we only need to consider up to <math>7 \mod 13</math> and <math>18 \mod 37</math>:
 +
 
 +
m |<math>m^4 \mod 37</math>| <math>m^4 \mod 13</math>
 +
0 |    0      |      0
 +
1 |    1      |      1
 +
2 |    16    |      3
 +
3 |    7      |      3
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4 |    34    |      9
 +
5 |    33    |      1
 +
6 |    1      |      9
 +
7 |    33    |      9
 +
8 |    26
 +
9 |    12
 +
10|    10
 +
11|    26
 +
12|    16
 +
13|    34
 +
14|    10
 +
15|    9
 +
16|    9
 +
17|    12
 +
18|    7
 +
 
 +
Note that the purpose of this table is to find an two <math>m^4</math> and <math>n^4</math> such that they add up to <math>0</math> for both <math>\mod 13</math> and <math>\mod 37</math>. Looking at <math>\mod 13</math>, this is clearly impossible unless both <math>m^4</math> and <math>n^4</math> are congruent to <math>0 \mod 13</math>. Thus, <math>m \equiv n \equiv 0 \mod 13</math>. We do the same for <math>37</math>, and after a bit of calculation, we see that it is also impossible for <math>m^4</math> and <math>n^4</math> to be anything other than <math>0 \mod 37</math>. Thus, <math>m</math> and <math>n</math> are both divisible by <math>13 \cdot 37</math>, or <math>481</math>. Logically, the first pair we test is <math>m = 2\cdot 13\cdot 37</math>, and <math>n = 13\cdot 37</math>. Indeed, this pair works with <math>a = 13\cdot 37\cdot 76</math> and <math>b = 13\cdot 37\cdot 49</math>. Since we know that <math>m > n</math> with <math>m \equiv n \equiv 0 \mod 481</math>, we know that this is the smallest possible value to work. Thus, the minimum of <math>{m^2, n^2}</math> is <math>481^2</math>, or <math>230400 + 961 = \boxed{231361}</math>.
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~Stead (first IMO problem solution! :D)
  
 
==Video Solution==
 
==Video Solution==
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==See Also==
 
==See Also==
  
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{{IMO box|year=1996|num-b=4|num-a=5}}
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{{IMO box|year=1996|num-b=3|num-a=5}}
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[[Category:Olympiad Geometry Problems]]
 
−
[[Category:3D Geometry Problems]]
 

Latest revision as of 10:00, 16 March 2025

Problem

The positive integers $a$ and $b$ are such that the numbers $15a+16b$ and $16a-15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?

Solution

Let us first set up two equations given the information in the problem: Let $15a + 16b = m^2$ Let $16a - 15b = n^2$

In general, when we are given alternating sums and differences like this, it is a good idea to square both equations and add them up to cancel out any cross terms. Doing this results in the equation $m^4 + n^4 = 481(a^2 + b^2)$.

Note that $481$ factorizes into $13 \cdot 37$. Thus, $m^4 + n^4 \equiv 0\mod 13$, and $m^4 + n^4 \equiv 0\mod 37$. Now, we make a table for all modular congruences possible for m^4 for all nonnegative integers with a unique modulo 13 and 37. Note that due to the modular symmetry of even powers ($(-a)^2 = a^2$), we only need to consider up to $7 \mod 13$ and $18 \mod 37$: 

m |$m^4 \mod 37$| $m^4 \mod 13$
0 |     0      |      0
1 |     1      |      1
2 |     16     |      3
3 |     7      |      3
4 |     34     |      9
5 |     33     |      1
6 |     1      |      9
7 |     33     |      9
8 |     26
9 |     12
10|     10
11|     26
12|     16
13|     34
14|     10
15|     9
16|     9
17|     12
18|     7

Note that the purpose of this table is to find an two $m^4$ and $n^4$ such that they add up to $0$ for both $\mod 13$ and $\mod 37$. Looking at $\mod 13$, this is clearly impossible unless both $m^4$ and $n^4$ are congruent to $0 \mod 13$. Thus, $m \equiv n \equiv 0 \mod 13$. We do the same for $37$, and after a bit of calculation, we see that it is also impossible for $m^4$ and $n^4$ to be anything other than $0 \mod 37$. Thus, $m$ and $n$ are both divisible by $13 \cdot 37$, or $481$. Logically, the first pair we test is $m = 2\cdot 13\cdot 37$, and $n = 13\cdot 37$. Indeed, this pair works with $a = 13\cdot 37\cdot 76$ and $b = 13\cdot 37\cdot 49$. Since we know that $m > n$ with $m \equiv n \equiv 0 \mod 481$, we know that this is the smallest possible value to work. Thus, the minimum of ${m^2, n^2}$ is $481^2$, or $230400 + 961 = \boxed{231361}$.

~Stead (first IMO problem solution! :D)

Video Solution

https://www.youtube.com/watch?v=d3Olg8LekzA&t=5s

See Also

1996 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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