Sum of squares to 30000?

Let a , b , c , d . . . a, b, c, d... be positive integers.

Is it possible to find such a set of integers such that a + b + c + . . . a + b + c + ... = 200 200 and a 2 + b 2 + c 2 + . . . a^2 + b^2 + c^2 + ... = 30000 30000 ?

Cannot be determined No Yes

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2 solutions

Denton Young
Aug 25, 2019

173 + 4 + 4 + 4 + 3 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 200 173 + 4 + 4 + 4 + 3 + 2 + 1 + 1 +1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 200

17 3 2 + 4 2 + 4 2 + 4 2 + 3 2 + 2 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 = 30000 173^2 + 4^2 + 4^2 + 4^2 + 3^2 + 2^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 +1^2 = 30000

Alternatively, 173 + 7 + 2 + 18 × 1 = 200 173+7+2+18\times 1=200 and 17 3 2 + 7 2 + 2 2 + 18 × 1 2 = 30000 173^2+7^2+2^2+18\times 1^2=30000 . I wonder how many solutions there are. I initially tried a theoretical approach but didn't really get anywhere - do you have any insights about when the answer to the general problem is "yes" and when it's "no"?

Chris Lewis - 1 year, 9 months ago

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Third solution: {172, 20, 2, 2, 2, 2}

I don't have any insights about the general problem.

Denton Young - 1 year, 9 months ago
Jon Haussmann
Sep 4, 2019

172 + 20 + 2 + 2 + 2 + 2 = 200 17 2 2 + 2 0 2 + 2 2 + 2 2 + 2 2 + 2 2 = 30000 \begin{aligned} 172 + 20 + 2 + 2 + 2 + 2 &= 200 \\ 172^2 + 20^2 + 2^2 + 2^2 + 2^2 + 2^2 &= 30000 \end{aligned}

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