System of quadratic homogeneous Diophantine equations

Question

Is there a method for determining if a system of quadratic diophantine equations has any solutions?

My specific question is: \begin{align*} a^2+b^2&=c^2\\[4pt] 2a^2+b^2&=d^2\\ \end{align*} I want to know if there are any positive integers $a, b, c, d$ which satisfy both equations.

The question came from my student and I cannot answer her in class.

Answer 1

For the system of equations:

$$\left\{\begin{aligned}&a^2+b^2=c^2\\&a^2+qb^2=w^2\end{aligned}\right.$$

If you can decompose the coefficient multipliers as follows: $q=(p\pm1)(s\pm1)$

Their work squares: $ps=t^2$

Then decisions can be recorded.

$$a=p-s$$

$$b=2t$$

$$c=p+s$$

$$w=\mp2q+p+s\pm2$$

You can add another simple option.

If the ratio can be written as: $q=2t^2-1$

Then decisions can be recorded.

$$a=t^2-1$$

$$b=2t$$

$$c=t^2+1$$

$$w=3t^2-1$$

Answer 2

Suppose $(a,b,c,d)$ is a solution to the system $$ \left\lbrace \begin{align*} &a^2+b^2=c^2\\[4pt] &2a^2+b^2=d^2\\[4pt] \end{align*} \right. $$ where $a,b,c,d$ are positive integers, and where $d$ is as small as possible.

If $a,b$ have a common factor $g > 1$, then $c$ and $d$ must also be divisible by $g$, hence $$\left(\frac{a}{g},\frac{b}{g},\frac{c}{g},\frac{d}{g}\right)$$ would also be a solution, contrary to the assumed minimality of $d$.

Hence $\gcd(a,b)=1$.

Suppose $b$ is even.

Then from $2a^2+b^2=d^2$, it follows that $d$ is even.

But then from $2a^2=d^2-b^2$, it follows that $2a^2$ is a multiple of $4$, hence $a$ is even, contrary to $\gcd(a,b)=1$.

Hence $b$ is odd.

Since $\gcd(a,b)=1$ and $b$ is odd, using the standard form for primitive Pythagorean triples, it follows that the equation $$a^2+b^2=c^2$$ holds if and only if $$ \left\lbrace \begin{align*} a&=2st\\[4pt] b&=s^2-t^2\\[4pt] c&=s^2+t^2\\[4pt] \end{align*} \right. $$ for some positive integers $s,t$ such that

• $s > t$.$\\[4pt]$
• $\gcd(s,t)=1$.$\\[4pt]$
• One of $s,t$ is even (and the other is odd).

Then the equation $2a^2+b^2=d^2$ can be recast as $$ 2(2st)^2+(s^2-t^2)^2=d^2 $$ or equivalently $$ s^4+6s^2t^2+t^4=d^2 $$ hence, letting $x={\Large{\frac{t}{s}}}$ and $y={\Large{\frac{d}{s}}}$, it follows that the equation $$y^2=x^4+6x^2+1$$ has a rational point $(x,y)$ with $0 < x < 1$.

That's as far as I can go.

I suspect there is no such rational point, but if so, I doubt that it can be proved in an elementary way.

Using more advanced methods, perhaps there is a way to convert the equation $y^2=x^4+6x^2+1$ to the equation of an elliptic curve in such a way that the non-existence of qualifying rational points on the curve $y^2=x^4+6x^2+1$ can be inferred from the non-existence of qualifying rational points on the elliptic curve.

Answer 3

$$y^2 = x^4+6x^2+1\tag{1}$$ Quartic equation $(1)$ can be transformed to elliptic curve below. $$Y^2 = X^3+6X^2+X\tag{2}$$ with $X = x^2.$

Using online Magma calculator as follows.
E:=EllipticCurve([$0, 6, 0, 1, 0$]);
IntegralPoints(E);
Rank(E);

It says that rank is $0$ and all integral points are $ [ (-1 : -2 : 1), (0 : 0 : 1) ]$.
Magma says that rank is $0$, and rank is $0$ means there is no rational point.

Hence only integral point is $(X,Y)=(0,0).$
From $x^2=X$, we get $(x,y)=(0,1).$
According to quasi's answer, since x=t/s,y=d/s then we get $(d,s,t)=(1,1,0).$
Finally, from $a=2st,b=s^2-t^2,c=s^2+t^2$, we get $(a,b,c,d)=(0,1,1,1).$