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| − | ==Problem==
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| − | Let <math>a,b</math> be two natural numbers. When we divide <math>a^2+b^2</math> by <math>a+b</math>, we the the remainder <math>r</math> and the quotient <math>q.</math> Determine all pairs <math>(a, b)</math> for which <math>q^2 + r = 1977.</math>
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| − | ==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>. If we now assume Wlog that <math>a\ge b</math>, 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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| − | The above solution was posted and copyrighted by cobbler. The original thread for this problem can be found here: [https://aops.com/community/p3404470]
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| − | == See Also == {{IMO box|year=1977|num-b=3|num-a=5}}
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