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Greatest integer and factorial part equation

Calculation of Real \((x,y,z)\)$ in

\(x[x]+z\{z\}-y\{y\} = 0.16\)

\(y[y]+x\{x\}-z\{z\} = 0.25\)

\(z[z]+y\{y\}-x\{x\} = 0.49\)

where \([x] = \)Greatest Integer of \(x\) and \(\{x\} = \) fractional part of \(x\)

Note by Jagdish Singh
3 years, 11 months ago

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(POST#2) - In post #1, we have discovered that one and only one of \([x]\), \([y]\) and \([z]\) must be \(1\) or \(-1\).


  • Case 1: \([x]=1\),\([y]=0\),\([z]=0\)
  • (1): \(x+z^2-y^2=0.16\) ---(4)
  • (2): \(x(x-1)-z^2=0.25\) ---(5)
  • (3): \(y^2-x(x-1)=0.49\) ---(6)
  • (4)+(5)+(6): \(x=0.9\), which contradicts the prerequisite.

  • Case 2: \([x]=-1\),\([y]=0\),\([z]=0\)
  • (1): \(-x+z^2-y^2=0.16\) ---(4)
  • (2): \(x(x+1)-z^2=0.25\) ---(5)
  • (3): \(y^2-x(x+1)=0.49\) ---(6)
  • (4)+(5)+(6): \(x=-0.9\) ---(7)
  • (7) into (6): \(y=\sqrt{0.4}\)
  • (7) into (5): \(z=\sqrt{-0.34}\), which is not real.

  • Case 3: \([x]=0\),\([y]=1\),\([z]=0\)
  • (1): \(z^2-y(y-1)=0.16\) ---(4)
  • (2): \(y+x^2-z^2=0.25\) ---(5)
  • (3): \(y(y-1)-x^2)=0.49\) ---(6)
  • (4)+(5)+(6): \(y=0.9\), which contradicts the prerequisite.
  • In fact, I expect case 5 to fail as well, because it would yield \(z=0.9\).
  • Therefore, I will skip case 5.

  • Case 4: \([x]=0\),\([y]=-1\),\([z]=0\)
  • (1): \(z^2-y(y+1)=0.16\) ---(4)
  • (2): \(-y+x^2-z^2=0.25\) ---(5)
  • (3): \(y(y+1)-x^2=0.49\) ---(6)
  • (4)+(5)+(6): \(y=-0.9\) ---(7)
  • (7) into (6): \(x=\pm\sqrt{-0.56}\), which is not real.

  • Case 5: \([x]=0\),\([y]=0\),\([z]=1\) (skipped)

  • Case 6: \([x]=0\),\([y]=0\),\([z]=-1\)
  • (1): \(z(z+1)+y^2=0.16\) ---(4)
  • (2): \(x^2-z(z+1)=0.25\) ---(5)
  • (3): \(-z+y^2-x^2)=0.49\) ---(6)
  • (4)+(5)+(6): \(z=-0.9\) ---(7)
  • (7) into (5): \(x=\sqrt{0.34}\)
  • (7) into (4): \(y=\sqrt{0.25}\)

  • \(x=\sqrt{0.34}, y=0.5, z=-0.9\).

Kenny Lau - 2 years, 11 months ago

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(POST#1)

  • Let \([x],\{x\},[y],\{y\},[z],\{z\}\) be \(a,b,c,d,e,f\) respectively.
  • \(\left\{\begin{array}{lclr}a^2+ab+ef+f^2-cd-d^2&=&0.16&\mbox{---(1)}\\c^2+cd+ab+b^2-ef-f^2&=&0.25&\mbox{---(2)}\\e^2+ef+cd+d^2-ab-b^2&=&0.49&\mbox{---(3)}\end{array}\right.\)
  • (1)+(2)+(3): \(a^2+ab+c^2+cd+e^2+ef = 0.9 \mbox{---(4)}\)
  • (4)-(2): \(a^2-b^2+(e+f)^2=0.65 \mbox{---(5)}\)
  • (4)-(3): \(c^2-d^2+(a+b)^2=0.41 \mbox{---(6)}\)
  • (4)-(1): \(e^2-f^2+(c+d)^2=0.74 \mbox{---(7)}\)
  • Since \(0\le b^2,d^2,f^2<1\),
  • (5): \(a^2+(e+f)^2=0.65+b^2<1.65\) ---(8)
  • (6): \(c^2+(a+b)^2=0.41+d^2<1.41\) ---(9)
  • (7): \(e^2+(c+d)^2=0.74+f^2<1.74\) ---(10)
  • Since \(0\le b,d,f<1\),
  • (8): \(a^2+e^2<1.65\) ---(11)
  • (9): \(c^2+a^2<1.41\) ---(12)
  • (10): \(e^2+c^2<1.74\) ---(13)
  • Since \(a\), \(c\), and \(e\) are integers, they can only be \(0\) or \(\pm1\) according to (11), (12) and (13).
  • Moreover, there can be at most one of \(a\), \(c\), or \(e\) which is \(\pm1\), or else (11), (12) or (13) will not be satisfied.
  • However, of all three of them are zero, (4) will not be satisfied.
  • In the next post, I shall explore all the three cases.

Kenny Lau - 2 years, 11 months ago

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