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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. · 3 years, 9 months ago

Wow..thanks a lot. · 3 years, 9 months ago

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?? · 3 years, 9 months ago

Not really. There are lots of options. I could have chosen $$0,9,18,27,\ldots$$, but I wanted small numbers. · 3 years, 9 months ago

so you arithmetically solved it on a piece of paper · 3 years, 9 months ago

Well, I played with the values of squares modulo $$9$$. · 3 years, 9 months ago

Thank you....I realized it later.... · 3 years, 9 months ago

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