Infinintely many squares of a form

[mutasimmim]

Does there exist infinite many pairs $(m,n)$ of positive integers such that $(a^2+b^2+3ab)$ is a perfect square?

[mutasimmim]

Solution:

$(3,7)$ is a solution. And so is $(3u , 7u)$

[SANZEED]

Does there exist infinite many pairs $(m,n)$ of positive integers such that $(a^2+b^2+3ab)$ is a perfect square?

What if we are asked to find all ordered $(a,b)$ such that $(a^2+b^2+3ab)$ is a perfect square?

[mutasimmim]

Then we will have to find all ordered pairs!

[Individ]

This equation is symmetric so many solutions. Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical. Write the formula can someone come in handy.

the equation: $Y^2+aXY+X^2=Z^2$

Has a solution:

$X=as^2-2ps$

$Y=p^2-s^2$

$Z=p^2-aps+s^2$

more:

$X=(4a+3a^2)s^2-2(2+a)ps-p^2$

$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$

$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$

more:

$X=(a+4)p^2-2ps$

$Y=3p^2-4ps+s^2$

$Z=(2a+5)p^2-(a+4)ps+s^2$

more:

$X=8s^2-4ps$

$Y=p^2-(4-2a)ps+a(a-4)s^2$

$Z=-p^2+4ps+(a^2-8)s^2$

For the particular case: $Y^2+XY+X^2=Z^2$

You can draw more formulas.

$X=3s^2+2ps$

$Y=p^2+2ps$

$Z=p^2+3ps+3s^2$

more:

$X=3s^2+2ps-p^2$

$Y=p^2+2ps-3s^2$

$Z=p^2+3s^2$

In the equation: $X^2+aXY+bY^2=Z^2$

there is always a solution and one of them is quite simple.

$X=s^2-bp^2$

$Y=ap^2+2ps$

$Z=bp^2+aps+s^2$

$p,s$ - integers asked us.

[*Mahi*]

Would you explain your methods of deriving these formulas? People of this forum would benefit highly from that.
I hope your answer is not Wolfram Alpha, seriously.

[Individ]

No. It's not Wolfram Alpha. He is such equations to solve not know how!
Although it is necessary to write the equations in a more General way.

$$aX^2+bXY+cY^2=jZ^2$$

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.

$$X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+$$

$$+(d+t-2c\pm\sqrt{j})s^2$$

$$Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2+2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+$$

$$+(d+t-2c\pm\sqrt{j})s^2$$

$$Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+$$

$$+(2(a+b-q)-d-t\pm\sqrt{j})s^2$$

$p,s$ are integers and are given us. 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.

The formula I gave in the first post - are derived from these.

[*Mahi*]

That is really great Thanks!

That you took your time to write this post in details is much, much appreciated.