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individ
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The problem is that when I got the formula that led below - there was a question whether they really describe all the decisions?
But has not yet found a counterexample, and it exists at all interested or not?
Although the discussion of this issue in many other forums and did not work. Everyone really does not like a lot of formula.
Here are the formulas. Formula generally look like this: Do not like these formulas. But this does not mean that we should not draw them. To start this equation zayimemsya, well then, and others.
$$aX^2+bXY+cY^2=jZ^2$$
Solutions can be written if even a single root. $$\sqrt{j(a+b+c)}$$ , $$\sqrt{b^2 + 4a(j-c)}$$ , $$\sqrt{b^2+4c(j-a)}$$
Then the solution can be written.
$$X=(2j(b+2c)^2-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+ +(j\mp \sqrt{j(a+b+c)})p^2$$
$$Y=(2j(2j-b-2a)(b+2c)-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+ +2((2j-2a-b)\sqrt{j(a+b+c)}\mp{j(b+2c)})sp+(j\mp\sqrt{j(a+b+c)})p^2$$
$$Z=(2j(b+2c)^2-(b^2+4c(j-a))(a+b+c\pm\sqrt{j(a+b+c)}))s^2+2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+( a + b + c \mp \sqrt{j(a+b+c)})p^2$$
In the case when the root $$\sqrt{b^2+4c(j-a)}$$ whole.
Solutions have the form.
$$X=((2j-b-2c)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+ +2(4ac+b(2j-b)\pm{(2j-b-2c)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Y=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(2j-b-2a\mp\sqrt{b^2+4c(j-a)}))s^2+ +2(4ac+b(2j-b)\pm{(b+2a)}\sqrt{b^2+4c(j-a)})sp+(2j-b-2a\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Z=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+ +2(4ac+b(2j-b)\pm {(b+2a)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
In the case when the root $$\sqrt{b^2+4a(j-c)}$$ whole.
Solutions have the form.
$$X=(2j^2(b+2a)-j(a+b+c)(2j-2c-b\pm\sqrt{b^2+4a(j-c)}))p^2+2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps++(2j-2c-b\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Y=(2j^2(b+2a)-j(a+b+c)(b+2a\pm\sqrt{b^2+4a(j-c)}))p^2 +2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+ +(b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Z=j(a+b+c)(b+2a\mp\sqrt{b^2+4a(j-c)})p^2+2((a+b+c)\sqrt{b^2+4a(j-c)}\mp{j(b+2a)})ps+ + (b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
Since these formulas are written in general terms, require a certain specificity calculations.If, after a permutation of the coefficients, no root is not an integer. You need to check whether there is an equivalent quadratic form in which, at least one root of a whole. Is usually sufficient to make the substitution $$X\longrightarrow{X+kY}$$ or more $$Y\longrightarrow{Y+kX}$$ In fact, this reduces to determining the existence of solutions in certain Pell's equation. Of course with such an idea can solve more complex equations. If I will not disturb anybody, slowly formula will draw. number $$p,s$$ integers and set us. I understand that these formulas do not like. And when they draw - or try to ignore or delete.
Formulas but there are no bad or good. They either are or they are not.
In equation $$aX^2+bY^2+cZ^2=qXY+dXZ+tYZ$$
$$a,b,c,q,d,t$$ integer coefficients which specify the conditions of the problem.
For a more compact notation, we introduce a replacement.
$$k=(q+t)^2-4b(a+c-d)$$
$$j=(d+t)^2-4c(a+b-q)$$
$$n=t(2a-t-d-q)+(2b-q)(2c-d)$$
Then the formula in the general form is:
$$X=(2n(2c-d-t)+j(q+t-2b\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2$$
$$Y=(2n(2c-d-t)+j(2(a+c-d)-q-t\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{ n })ps+ +(q+t+2(d-a-c)\pm\sqrt{k})s^2$$
$$Z=(j(q+t-2b\pm\sqrt{k})-2n(2(a+b-q)-d-t))p^2+2((2(a+b-q)-d-t)\sqrt{k}\mp{n})ps+ +(2b-q-t\pm\sqrt{k})s^2$$
And more.
$$X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2$$
$$Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2+2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+ +(d+t-2c\pm\sqrt{j})s^2$$
$$Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+ +(2(a+b-q)-d-t\pm\sqrt{j})s^2$$
Since formulas are written in general terms, in the case where neither the root is not an integer, it is necessary to check whether there is such an equivalent quadratic form in which at least one root of a whole. If not, then the solution in integers of the equation have not.
But has not yet found a counterexample, and it exists at all interested or not?
Although the discussion of this issue in many other forums and did not work. Everyone really does not like a lot of formula.
Here are the formulas. Formula generally look like this: Do not like these formulas. But this does not mean that we should not draw them. To start this equation zayimemsya, well then, and others.
$$aX^2+bXY+cY^2=jZ^2$$
Solutions can be written if even a single root. $$\sqrt{j(a+b+c)}$$ , $$\sqrt{b^2 + 4a(j-c)}$$ , $$\sqrt{b^2+4c(j-a)}$$
Then the solution can be written.
$$X=(2j(b+2c)^2-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+ +(j\mp \sqrt{j(a+b+c)})p^2$$
$$Y=(2j(2j-b-2a)(b+2c)-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+ +2((2j-2a-b)\sqrt{j(a+b+c)}\mp{j(b+2c)})sp+(j\mp\sqrt{j(a+b+c)})p^2$$
$$Z=(2j(b+2c)^2-(b^2+4c(j-a))(a+b+c\pm\sqrt{j(a+b+c)}))s^2+2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+( a + b + c \mp \sqrt{j(a+b+c)})p^2$$
In the case when the root $$\sqrt{b^2+4c(j-a)}$$ whole.
Solutions have the form.
$$X=((2j-b-2c)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+ +2(4ac+b(2j-b)\pm{(2j-b-2c)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Y=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(2j-b-2a\mp\sqrt{b^2+4c(j-a)}))s^2+ +2(4ac+b(2j-b)\pm{(b+2a)}\sqrt{b^2+4c(j-a)})sp+(2j-b-2a\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Z=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+ +2(4ac+b(2j-b)\pm {(b+2a)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
In the case when the root $$\sqrt{b^2+4a(j-c)}$$ whole.
Solutions have the form.
$$X=(2j^2(b+2a)-j(a+b+c)(2j-2c-b\pm\sqrt{b^2+4a(j-c)}))p^2+2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps++(2j-2c-b\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Y=(2j^2(b+2a)-j(a+b+c)(b+2a\pm\sqrt{b^2+4a(j-c)}))p^2 +2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+ +(b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Z=j(a+b+c)(b+2a\mp\sqrt{b^2+4a(j-c)})p^2+2((a+b+c)\sqrt{b^2+4a(j-c)}\mp{j(b+2a)})ps+ + (b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
Since these formulas are written in general terms, require a certain specificity calculations.If, after a permutation of the coefficients, no root is not an integer. You need to check whether there is an equivalent quadratic form in which, at least one root of a whole. Is usually sufficient to make the substitution $$X\longrightarrow{X+kY}$$ or more $$Y\longrightarrow{Y+kX}$$ In fact, this reduces to determining the existence of solutions in certain Pell's equation. Of course with such an idea can solve more complex equations. If I will not disturb anybody, slowly formula will draw. number $$p,s$$ integers and set us. I understand that these formulas do not like. And when they draw - or try to ignore or delete.
Formulas but there are no bad or good. They either are or they are not.
In equation $$aX^2+bY^2+cZ^2=qXY+dXZ+tYZ$$
$$a,b,c,q,d,t$$ integer coefficients which specify the conditions of the problem.
For a more compact notation, we introduce a replacement.
$$k=(q+t)^2-4b(a+c-d)$$
$$j=(d+t)^2-4c(a+b-q)$$
$$n=t(2a-t-d-q)+(2b-q)(2c-d)$$
Then the formula in the general form is:
$$X=(2n(2c-d-t)+j(q+t-2b\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2$$
$$Y=(2n(2c-d-t)+j(2(a+c-d)-q-t\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{ n })ps+ +(q+t+2(d-a-c)\pm\sqrt{k})s^2$$
$$Z=(j(q+t-2b\pm\sqrt{k})-2n(2(a+b-q)-d-t))p^2+2((2(a+b-q)-d-t)\sqrt{k}\mp{n})ps+ +(2b-q-t\pm\sqrt{k})s^2$$
And more.
$$X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2$$
$$Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2+2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+ +(d+t-2c\pm\sqrt{j})s^2$$
$$Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+ +(2(a+b-q)-d-t\pm\sqrt{j})s^2$$
Since formulas are written in general terms, in the case where neither the root is not an integer, it is necessary to check whether there is such an equivalent quadratic form in which at least one root of a whole. If not, then the solution in integers of the equation have not.
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