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Solve in natural numbers

  • Thread starter oszust001
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  • #1
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How can I solve that type of equation:
[tex]x^2+y^2=4z^2[/tex] or [tex]x^2+3y^2=4z^2[/tex]
 

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  • #2
Char. Limit
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Both of those, so far as I know, have infinite solutions.
 
  • #3
HallsofIvy
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How can I solve that type of equation:
[tex]x^2+y^2=4z^2[/tex] or [tex]x^2+3y^2=4z^2[/tex]
Given any values of x and y, we could then solve for z so these have an infinite number of solutions. Since you are asking for solutions in natural numbers, you could treat them as variations on the Pythagorean theorem and use the formulas for producing integer solutions to [itex]x^2+ y^2= z^2[/itex]: given any two integers, m and n, then [itex]x= m^2- n^2[/itex], [itex]y= 2mn[/itex], and [itex]z= m^2+ n^2[/itex], satisfy [itex]x^2+ y^2= z^2[/itex]. You would need to exempt some values of m and n to allow for the "3" and "4" coefficients.
 
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