Solve in integers: $(x + y + z + v)^2 - 29w = 2023, xy + yz + zv + vx - w^2 = 475$

Question

Problem: Solve in integers the system of equations: \begin{align} (x + y + z + v)^2 - 29w &= 2023, \\ % Equation 1 xy + yz + zv + vx - w^2 &= 475. % Equation 2 \end{align}

This problem is from the algebra section of a local high school math competition.

My Work

Let's say $S = x + y + z + v$. We have:

$S^2 - 29w = 2023$

$xy + yz + zv + vx - w^2 = 475$

The second equation is messy, but I will try to rewrite it using $S$.

$2(xy + yz + zv + vx) = S^2 - (x^2 + y^2 + z^2 + v^2)$

Substitute this into the second equation:

$S^2 - (x^2 + y^2 + z^2 + v^2) - 2w^2 = 950$

So, $x^2 + y^2 + z^2 + v^2 + w^2 = S^2 - 950$

From here, I know that I can substitute $S^2$ back into the first equation, but I do not think that I will get anywhere with that. Could I please get some help with where to go from here or if I have completely taken the wrong path? Does the solution have to do with Diophantine Equations? Thank you for your help!

Answer 1

Substitute $x+z=a, y+v=b$. Now the system turns into:

$$(a+b)^2-29w=2023$$ $$ab-w^2=475$$

In other words, $2023+29w=(a+b)^2\ge 4ab=4(475+w^2)=4w^2+1900$ (using a well-known inequality $(a+b)^2\ge 4ab$).

So, we are getting $4w^2-29w-123\le 0$, which has solutions $w\in[-3, 10.25]$, so as $w$ is an integer, it is in the range $-3,\ldots, 10$.

Using the first equation, we can quickly sieve out some values of $w$, because $2023+29w=(a+b)^2$ must be a square number. Of course, if $-3\le w\le 10$, then $1936\le 2023+29w\le 2313$, and in this range the only square numbers are $44^2=1936, 45^2=2025, 46^2=2116, 47^2=2209, 48^2=2304$. Out of those, only $1936=2023-3\cdot 29$ gives an integer value $w=-3$ and this gives $a+b=\pm 44$.

Now, we can find $ab=475+w^2=484$ and then solve for $a, b$ using Vieta formulae i.e. solving the quadratic equation $X^2\mp 44X+484=0$. This gives $a=b=\pm 22$.

Finally, the solutions are as follows: $x, y$ can be arbitrary integers, $w=-3$; for the rest, either $z=22-x, v=22-y$ or $z=-22-x, v=-22-y$.

Answer 2

In this task, we are given two equations and we need to solve for integers. The given equations are:

1. (x+y+z+v)^2 − 29w = 2023
2. xy+yz+zv+vx − w^2 = 475

To solve the given equations, we need to use algebraic manipulation and solve for each variable step by step. Let's begin with the first equation:

1. (x+y+z+v)^2 − 29w = 2023

Expanding the left side of the equation, we get:

(x^2 + y^2 + z^2 + v^2 + 2xy + 2xz + 2xv + 2yz + 2yv + 2zv) − 29w = 2023

Now, we can simplify the equation by grouping the variables:

(x^2 + y^2 + z^2 + v^2) + 2(xy + xz + xv + yz + yv + zv) − 29w = 2023

Let's focus on the second term and factor it:

2(xy + xz + xv + yz + yv + zv) = 2[(x+y)(z+v) + xz + yv]

Now, we can substitute this back into the equation and simplify further:

(x^2 + y^2 + z^2 + v^2) + 2[(x+y)(z+v) + xz + yv] − 29w = 2023 x^2 + y^2 + z^2 + v^2 + 2(x+y)(z+v) + 2xz + 2yv − 29w = 2023

At this point, we can see that we have four variables and only one equation. We need another equation to solve for these variables. Fortunately, we have another equation given to us. Let's use it:

2. xy+yz+zv+vx − w^2 = 475

We can rearrange this equation to get:

w^2 = xy+yz+zv+vx − 475

Now, we can substitute this back into the first equation:

x^2 + y^2 + z^2 + v^2 + 2(x+y)(z+v) + 2xz + 2yv − 29(xy+yz+zv+vx − 475) = 2023

Expanding and simplifying, we get:

x^2 + y^2 + z^2 + v^2 − 27xy − 27yz − 27zv − 27vx + 58xz + 58yv + 29w = 13848

Now, we can substitute the value of w^2 from the second equation:

x^2 + y^2 + z^2 + v^2 − 27xy − 27yz − 27zv − 27vx + 58xz + 58yv + 29(xy+yz+zv+vx − 475) = 13848

Simplifying further, we get:

x^2 + y^2 + z^2 + v^2 − 1555 = 27(xy + yz + zv + vx − 475)

We can simplify the right side of the equation further by using the value of w^2:

x^2 + y^2 + z^2 + v^2 − 1555 = 27(xy + yz + zv + vx) − 27w^2

Substituting the value of w^2, we get:

x^2 + y^2 + z^2 + v^2 − 1555 = 27(xy + yz + zv + vx) − 27(xy+yz+zv+vx − 475)

Simplifying, we get:

27w^2 = 13002

Now, we can solve for w:

w^2 = 482 w = ±22

Substituting w = 22 in the equation x^2 + y^2 + z^2 + v^2 − 1555 = 27(xy + yz + zv + vx) − 27w^2, we get:

x^2 + y^2 + z^2 + v^2 = 809 + 27(xy + yz + zv + vx)

We can see that the left side of the equation is a sum of squares, which means that the right side must also be a sum of squares. Therefore, we need to find four integers whose sum of squares is equal to 809. By trial and error, we can find that:

x = −13

y = −2

z =