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AMTI Bhaskara Contest Final Junior (October 2013) question no. 8(b)

Do there exist 10 distinct integers such that sum of any 9 of which is a perfect square?

Note by Piyushkumar Palan
3 years, 9 months ago

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Suppose that the numbers are \(x_1,x_2,\ldots,x_{10}\), and suppose that the sum of all the numbers except the \(j\)th is the square \(y_j^2\). If \(N = x_1+x_2+\cdots+x_{10}\), we deduce that \[ x_j \; = \; N - y_j^2 \qquad 1 \le j \le 10 \] For this to be consistent, we need \[ N \; = \; x_1 + x_2 + \cdots + x_{10} \; = \; 10N - (y_1^2 + y_2^2 + \cdots + y_{10}^2) \] so that \[ y_1^2 + y_2^2 + \cdots + y_{10}^2 \; = \; 9N \] We need to choose \(y_1,y_2,\cdots,y_{10}\) to be distinct integers such that the sum of their squares is divisible by \(9\), for example \(2,3,4,5,6,7,8,9,11,12\). This gives \(N=61\) and \[ x_1,x_2,\ldots,x_{10} \; = \; 57,52,45,36,25,12,-3,-20,-60,-83 \] There is no rule that says that the integers have to be positive. Mark Hennings · 3 years, 9 months ago

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@Mark Hennings Wow..thanks a lot. Piyushkumar Palan · 3 years, 9 months ago

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@Mark Hennings Ahhh....loved it....the method of solving was awesome.......but you did take time to solve the integers \(2,3,4,5,6,7,8,9,11,12\)....right? If not,could you please tell me how you found those integers?? Vaibhav Reddy · 3 years, 9 months ago

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@Vaibhav Reddy Not really. There are lots of options. I could have chosen \(0,9,18,27,\ldots\), but I wanted small numbers. Mark Hennings · 3 years, 9 months ago

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@Mark Hennings so you arithmetically solved it on a piece of paper Vaibhav Reddy · 3 years, 9 months ago

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@Vaibhav Reddy Well, I played with the values of squares modulo \(9\). Mark Hennings · 3 years, 9 months ago

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@Mark Hennings Thank you....I realized it later.... Vaibhav Reddy · 3 years, 9 months ago

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This may be simple or difficult ..i tried for a while...need help! Piyushkumar Palan · 3 years, 9 months ago

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Why graph of quadratic equation is parabola ? Vaibhav Gandhapwad · 2 years, 2 months ago

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