Find 7 positive integers

Find seven positive integers x 1 , x 2 , x 3 , . . . x 7 x_1, x_2, x_3, ...x_7 such that

( x 1 x 2 x 3 . . . x 7 ) 2 = 2 ( x 1 2 + x 2 2 + x 3 2 + . . . + x 7 2 ) \left(x_1x_2x_3...x_7\right)^2=2\left(x_1^2+x_2^2+x_3^2 + ...+x_7^2\right)

Type your answer as the sum of these seven positive integers. If you think that there is no correct answer, type 0 0 , or there are more than one distinct solution, type 1 -1 .

Clarification: Reordering the sequence of those integers does not count as another solution.


The answer is 10.

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1 solution

Chris Lewis
Jul 25, 2020

The only set that works (up to reordering) is { 1 , 1 , 1 , 1 , 1 , 2 , 3 } \{1,1,1,1,1,2,3\} .

Without loss of generality, say x 1 x 2 x 7 x_1 \le x_2 \le \cdots \le x_7 .

The right hand side of the equation is at most 14 x 7 2 14x_7^2 .

From the left hand side, this means x 1 x 2 x 3 x 4 x 5 x 6 3 x_1 x_2 x_3 x_4 x_5 x_6 \le 3 , which only leaves three possible sets to check: we must have x 1 = x 2 = x 3 = x 4 = x 5 = 1 x_1=x_2=x_3=x_4=x_5=1 , then x 6 x_6 is 1 , 2 1,2 or 3 3 .

As it happens, only the one set above works.

just amazing!

Alexander Shannon - 10 months, 2 weeks ago

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