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 A Cubic And Biquadratic Problem (Posted on 2006-06-18)
Find the three smallest positive integers (with the second number exceeding the first) such that the sum of the squares of the first two numbers is equal to the cube of the third number while the sum of the cubes of the first two numbers is equal to nine times the fourth power of the third number.

 See The Solution Submitted by K Sengupta Rating: 3.0000 (2 votes)

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 computer solution (spoiler) Comment 1 of 1

The below program was halted at a total of 24576 for the first two numbers, so the shown solution (625, 1250, 125) and pseudosolution (1458, 1458, 162, where B does not exceed A) are the only two that exist for the total of the first two to be less than 24,576.

`list   10   for T=1 to 1000000   20   for A=1 to int(T/2)   30    B=T-A   40    Cc=A*A+B*B   50    C=int(Cc^(1/3)+0.5)   60    if C*C*C=Cc then   70      :if A*A*A+B*B*B=9*Cc*C then   80      :print A,B,C   90   next A  100   next TOKrun 625     1250    125 1458    1458    162Break in 50asave "cubbiq"OKcontBreak in 50?t 7965OKcontBreak in 50?t 12650OKcontBreak in 50?t 24576OK`

 Posted by Charlie on 2006-06-18 13:22:25

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