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Difference between revisions of "1977 IMO Problems/Problem 5"

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==Solution==
 
==Solution==
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Using <math>r=1977-q^2</math>, we have <math>a^2+b^2=(a+b)q+1977-q^2</math>, or <math>q^2-(a+b)q+a^2+b^2-1977=0</math>, which implies <math>\Delta=7908+2ab-2(a^2+b^2)\ge 0</math>. Using AM-GM inequality and a,b>0, we have 3(a+b)^2<=7908+8ab<=7098+8(a+b)^2/4, it follows <math>a+b\le 88</math>. If <math>q\le 43</math>, then <math>r=1977-q^2\ge 128</math>, contradicting <math>r<a+b\le 88</math>. But <math>q\le 44</math> from <math>q^2+r=1977</math>, thus <math>q=44</math>. It follows <math>r=41</math>, and we get <math>a^2+b^2=44(a+b)+41\Leftrightarrow (a-22)^2+(b-22)^2=1009\in \mathbb{P}</math>. By Jacobi's two squares theorem, we infer that <math>15^2+28^2=1009</math> is the only representation of <math>1009</math> as a sum of squares. This forces <math>\boxed{(a,b)=(37,50) , (7, 50)}</math>, and permutations. <math>\blacksquare</math>
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Using <math>r=1977-q^2</math>, we have <math>a^2+b^2=(a+b)q+1977-q^2</math>, or <math>q^2-(a+b)q+a^2+b^2-1977=0</math>, which implies <math>\Delta=7908+2ab-2(a^2+b^2)\ge 0</math>. Using AM-GM inequality and the fact a,b>0, we have 3(a+b)^2<=7908+8ab<=7098+8(a+b)^2/4, it follows <math>a+b\le 88</math>. If <math>q\le 43</math>, then <math>r=1977-q^2\ge 128</math>, contradicting <math>r<a+b\le 88</math>. But <math>q\le 44</math> from <math>q^2+r=1977</math>, thus <math>q=44</math>. It follows <math>r=41</math>, and we get <math>a^2+b^2=44(a+b)+41\Leftrightarrow (a-22)^2+(b-22)^2=1009\in \mathbb{P}</math>. By Jacobi's two squares theorem, we infer that <math>15^2+28^2=1009</math> is the only representation of <math>1009</math> as a sum of squares. This forces <math>\boxed{(a,b)=(37,50) , (7, 50)}</math>, and permutations. <math>\blacksquare</math>
  
 
The above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here: [https://aops.com/community/p3404470]
 
The above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here: [https://aops.com/community/p3404470]
  
 
== See Also == {{IMO box|year=1977|num-b=4|num-a=6}}
 
== See Also == {{IMO box|year=1977|num-b=4|num-a=6}}

Latest revision as of 07:09, 21 January 2025

Problem

Let $a,b$ be two natural numbers. When we divide $a^2+b^2$ by $a+b$, we the the remainder $r$ and the quotient $q.$ Determine all pairs $(a, b)$ for which $q^2 + r = 1977.$

Solution

Using $r=1977-q^2$, we have $a^2+b^2=(a+b)q+1977-q^2$, or $q^2-(a+b)q+a^2+b^2-1977=0$, which implies $\Delta=7908+2ab-2(a^2+b^2)\ge 0$. Using AM-GM inequality and the fact a,b>0, we have 3(a+b)^2<=7908+8ab<=7098+8(a+b)^2/4, it follows $a+b\le 88$. If $q\le 43$, then $r=1977-q^2\ge 128$, contradicting $r<a+b\le 88$. But $q\le 44$ from $q^2+r=1977$, thus $q=44$. It follows $r=41$, and we get $a^2+b^2=44(a+b)+41\Leftrightarrow (a-22)^2+(b-22)^2=1009\in \mathbb{P}$. By Jacobi's two squares theorem, we infer that $15^2+28^2=1009$ is the only representation of $1009$ as a sum of squares. This forces $\boxed{(a,b)=(37,50) , (7, 50)}$, and permutations. $\blacksquare$

The above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here: [1]

See Also

1977 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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