Does $ x_1^2 + x_2^2 + x_3^3 + x_4^3 = y_1^2 + y_2^2 + y_3^3 + y_4^3 $ have infinitely many solutions?

There are, at least, infinitely many rational solutions. Suppose you have two rational solutions $$\mathbf s_1=(x^{(1)}_1,x^{(1)}_2,x^{(1)}_3,x^{(1)}_4,y^{(1)}_1,y^{(1)}_,y^{(1)}_3,y^{(1)}_4)$$ and $$\mathbf s_2=(x^{(2)}_1,x^{(2)}_2,x^{(2)}_3,x^{(2)}_4,y^{(2)}_1,y^{(2)}_,y^{(2)}_3,y^{(2)}_4),$$ which both satisfy $f(\mathbf s_i)=0$. Here, $$f(x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4)=x_1^2+x_2^2+x_3^3+x_4^3-y_1^2-y_2^2-y_3^3-y_4^3.$$

Consider $p(t) = f(t\mathbf s_1+(1-t)\mathbf s_2)$. This is a cubic polynomial in $t$ with rational coefficients. However, $p(0)=0$ and $p(1)=0$, so $p(t)$ factorises as $t(1-t)(qt-r)$ for some rational $q$ and $r$. This means $\frac rq\mathbf s_1+(1-\frac rq)\mathbf s_2$ is a new solution.

You can also build new solutions by changing the signs of $x_1$, $x_2$, $y_1$ and $y_2$, or by choosing certain permutations of the solution. These various constructions together can be applied ad libitum to create as many rational solutions as you want.

To complete this argument, we need to demonstrate the existence of $\mathbf s_1$ and $\mathbf s_2$, but you've probably already noticed there are many solutions. Here's a few examples:

• $2^2+7^2+2^3+2^3=69=4^2+5^2+1^3+3^3$
• $3^2+8^2+1^3+1^3=75=2^2+6^2+2^3+3^3$
• $5^2+5^2+1^3+1^3=52=1^2+4^2+2^3+3^3$

For example, if we choose $\mathbf s_1=(3,8,1,1,2,6,2,3)$ and $\mathbf s_2=(1,4,3,2,5,5,1,1)$, we get $p(t)=-18t^3+37t^2-19t=t(1-t)(18t-19)$. This gives a new solution $$\left(\frac{28}9\right)^2+\left(\frac{74}9\right)^2+\left(\frac{8}9\right)^3+\left(\frac{17}{18}\right)^3=\frac{17027}{216}= \left(\frac{11}6\right)^2+\left(\frac{109}{18}\right)^2+\left(\frac{37}{18}\right)^3+\left(\frac{28}9\right)^3$$

But perhaps you want integer solutions. Suppose we have a solution $$\left(\frac {p_1}q,\frac {p_2}q,\frac {p_3}q,\frac {p_4}q,\frac {p_5}q,\frac {p_6}q,\frac {p_7}q,\frac {p_8}q\right)$$ where the $p_i$ and $q$ are integers (note any rational solution can be written over a common denominator). Then $$\left(\frac {p_1}q\right)^2+\left(\frac {p_2}q\right)^2+\left(\frac {p_3}q\right)^3+\left(\frac {p_4}q\right)^3=\left(\frac {p_5}q\right)^2+\left(\frac {p_6}q\right)^2+\left(\frac {p_7}q\right)^3+\left(\frac {p_8}q\right)^3$$

Multiplying through by $q^6$ promotes this to a new solution in integers:

$$\left({p_1}q^2\right)^2+\left({p_2}q^2\right)^2+\left( {p_3}q\right)^3+\left({p_4}q\right)^3=\left( {p_5}q^2\right)^2+\left( {p_6}q^2\right)^2+\left({p_7}q\right)^3+\left({p_8}q\right)^3$$

For example,

$$\left(\frac{28}9\right)^2+\left(\frac{74}9\right)^2+\left(\frac{8}9\right)^3+\left(\frac{17}{18}\right)^3=\frac{17027}{216}= \left(\frac{11}6\right)^2+\left(\frac{109}{18}\right)^2+\left(\frac{37}{18}\right)^3+\left(\frac{28}9\right)^3$$

becomes

$$18144^2+47952^2+288^3+306^3=2681139528=10692^2+35316^2+666^3+1008^3$$

It is known that there are infinitely many (integer) solutions for $x^3+y^3=z^3+w^3$ and so is for $x^2+y^2=z^2+w^2$. Consequently so is for $x_1^2 + x_2^2 + x_3^3 + x_4^3 = y_1^2 + y_2^2 + y_3^3 + y_4^3$ because if $x^3+y^3=z^3+w^3=m$ and $x^2+y^2=z^2+w^2=n$ then $m+n$ gives a solution. For example using the identity $$(2a+b)^2+(2b-a)^2=(2a-b)^2+(a+2b)^2\text{ and }\space 10^3+27^3=19^3+24^3$$ we get $$10^3+27^3+(2a+b)^2+(2b-a)^2=19^3+24^3+(2a-b)^2+(a+2b)^2$$ which satisfies our purpose of an infinite set of solutions..

There will be no complicated explanation here, all the forms are of the form: $$f(r,...r_n)+X^2=Y^2+f(t..,t_n)$$ $$x_1^2+x_2^2+x_3^3+x_4^3=y_1^2+y_2^2+y_3^3+y_4^3$$ are always solvable by substitution $X=x+y, Y=x-y.$ we can apply a substitution preserving the equivalence of the set of rational solutions. $$ \begin{cases} x_1=a+b \\ y_1=a-b\end{cases} $$

$$4ab+x_2^2+x_3^3+x_4^3=y_2^2+y_3^3+y_4^3$$ $$a=\frac{y_2^2+y_3^3+y_4^3-x_4^3-x_3^3-x_2^2}{4b}$$ $$\tiny(\frac{y_2^2+y_3^3+y_4^3-x_4^3-x_3^3-x_2^2}{4b}+b)^2+x_2^2+x_3^3+x_4^3=(\frac{y_2^2+y_3^3+y_4^3-x_4^3-x_3^3-x_2^2}{4b}-b)^2+y_2^2+y_3^3+y_4^3$$ $$\small(2b^2(y_2^2+y_3^3+y_4^3-x_4^3-x_3^3-x_2^2)+8b^4)^2+(8b^3x_2)^2+(4b^2x_3)^3+(4b^2x_4)^3=(2b^2(y_2^2+y_3^3+y_4^3-x_4^3-x_3^3-x_2^2)-8b^4)^2+(8b^3y_2)^2+(4b^2y_3)^3+(4b^2y_4)^3$$

let's do the scaling, now we just need to find the primitives among which the numbers must have conditions $x_2,x_3,x_4,\in \mathbb{Z}$, $y_2,y_3,y_4,\in \mathbb{Z}$, $b\in \mathbb{R}$ then we get a general family for the whole equation

• Example $b=1, x_2=2, x_3=3, x_4=4, y_2=5, y_3=6, y_4=7$ we get an infinite set of primitives. $$986^2+16^2+12^3+16^3=970^2+40^2+24^3+28^3$$

Let $$N=y_3^3+y_4^3-x_3^3-x_4^3$$ where $N>0.$ We now assume $N$ is a multiple of $4.$ [no reason except it ends up working.] Note that because $(k+1)^2-(k-1)^2=4k,$ any multiple of $4$ is a difference of squares, allowing us to find values for the remaining variables to cause equality of N with $M=[x_1^2-y_1^2]+[x_2^2-y_2^2.]$

The point here is we may select the cubes almost at random and then fill out the squares to make the 8-variable equation hold.

Example: $N=1^3+7^3-3^3-5^3=192.$ Then express $192$ as a sum of two multiples of $4$, say as $192=4 \cdot 30 + 4 \cdot 18.$ These values lead to the solution $$31^2+19^2+3^3+5^3=29^2+17^2+1^3+7^3.$$