Kvant Math Problem 1118

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Problem

1. Prove that the equation $$(x+y+z)^2=x^2+y^2+z^2\tag{$*$}$$ has infinitely many integer solutions.
2. How many such solutions are there for which $z=1988$?

S. G. Mamikonyan

Exploration

Expanding the left-hand side gives

$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx).$$

Hence $(*)$ is equivalent to

$$xy+yz+zx=0.$$

The first question asks for infinitely many integer triples satisfying this quadratic equation. A natural way to proceed is to solve for one variable:

$$z(x+y)=-xy.$$

If $x+y\neq0$, then

$$z=-\frac{xy}{x+y}.$$

To obtain integers systematically, write $x=am,\ y=an$. Then

$$z=-\frac{a^2mn}{a(m+n)}=-\frac{amn}{m+n}.$$

We need $m+n\mid amn$. Choosing $a=m+n$ gives

$$x=m(m+n),\qquad y=n(m+n),\qquad z=-mn.$$

Substitution yields

$$xy+yz+zx =mn(m+n)^2-mn^2(m+n)-m^2n(m+n)=0.$$

Since $m,n$ are arbitrary integers, this produces infinitely many solutions.

For the second question, put $z=1988$. Then

$$xy+1988(x+y)=0.$$

Adding $1988^2$ to both sides gives

$$(x+1988)(y+1988)=1988^2.$$

Thus every integer divisor $d$ of $1988^2$ gives a solution

$$x=d-1988,\qquad y=\frac{1988^2}{d}-1988.$$

The number of solutions is therefore the number of integer divisors of $1988^2$.

Factor

$$1988=2^2\cdot7\cdot71,$$

hence

$$1988^2=2^4\cdot7^2\cdot71^2.$$

The number of positive divisors is

$$(4+1)(2+1)(2+1)=45.$$

Since $1988^2>0$, both factors $(x+1988)$ and $(y+1988)$ must have the same sign. Positive divisors contribute $45$ ordered pairs, and negative divisors contribute another $45$. Thus there are $90$ integer solutions.

The step most likely to hide an error is the counting. One must count ordered pairs $(x,y)$, not unordered pairs, and include both positive and negative divisors.

Problem Understanding

The equation

$$(x+y+z)^2=x^2+y^2+z^2$$

must be studied over the integers.

The problem has two parts. First, one must prove that infinitely many integer solutions exist. Second, one must count all integer solutions with the fixed value $z=1988$.

This is a Type D problem. We must construct infinitely many solutions and then determine exactly how many solutions occur when $z$ is prescribed.

The core difficulty is recognizing that the quadratic equation simplifies to

$$xy+yz+zx=0,$$

and that when $z$ is fixed, it can be transformed into a divisor-counting problem by completing a product.

Proof Architecture

The first claim is that $(*)$ is equivalent to $xy+yz+zx=0$; this follows by expanding the square.

The second claim is that for arbitrary integers $m,n$,

$$x=m(m+n),\quad y=n(m+n),\quad z=-mn$$

satisfy $xy+yz+zx=0$; direct substitution proves this.

The third claim is that these formulas yield infinitely many distinct integer solutions; taking $m=1$ and varying $n$ suffices.

The fourth claim is that when $z=1988$,

$$xy+1988(x+y)=0$$

is equivalent to

$$(x+1988)(y+1988)=1988^2;$$

this follows by expansion.

The fifth claim is that the number of integer solutions equals the number of integer divisors of $1988^2$; each divisor determines a unique ordered pair $(x,y)$.

The sixth claim is that $1988^2=2^4\cdot7^2\cdot71^2$ has $45$ positive divisors and $90$ integer divisors; this is the standard divisor-count formula.

The lemma most likely to fail under scrutiny is the counting of solutions from the divisor equation, because one must count ordered pairs and include negative divisors.

Solution

Expanding the left-hand side of $(*)$ gives

$$x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2.$$

Hence $(*)$ is equivalent to

$$xy+yz+zx=0. \tag{1}$$

For the first part, let $m$ and $n$ be arbitrary integers and define

$$x=m(m+n),\qquad y=n(m+n),\qquad z=-mn.$$

Then

$$xy=m n (m+n)^2,$$

$$yz=-m n^2(m+n),$$

and

$$zx=-m^2 n(m+n).$$

Therefore

$$xy+yz+zx =mn(m+n)\bigl((m+n)-n-m\bigr) =0.$$

Thus every pair of integers $m,n$ produces a solution of $(1)$, and hence of $(*)$.

Taking $m=1$ and allowing $n$ to vary through all integers yields the family

$$x=n+1,\qquad y=n(n+1),\qquad z=-n.$$

Different values of $n$ give different triples, so infinitely many integer solutions exist.

Now impose the condition $z=1988$. Equation $(1)$ becomes

$$xy+1988(x+y)=0.$$

Adding $1988^2$ to both sides yields

$$xy+1988x+1988y+1988^2=1988^2,$$

hence

$$(x+1988)(y+1988)=1988^2. \tag{2}$$

Conversely, every solution of $(2)$ satisfies the previous equation, so the two equations are equivalent.

Factor

$$1988=2^2\cdot7\cdot71,$$

therefore

$$1988^2=2^4\cdot7^2\cdot71^2.$$

The number of positive divisors of $1988^2$ is

$$(4+1)(2+1)(2+1)=45.$$

Since $1988^2>0$, every integer divisor $d$ of $1988^2$ determines exactly one ordered pair

$$x=d-1988,\qquad y=\frac{1988^2}{d}-1988.$$

Positive divisors give $45$ solutions, and negative divisors give another $45$. Thus the total number of integer divisors of $1988^2$ is

$$2\cdot45=90.$$

Hence there are exactly $90$ integer solutions with $z=1988$.

The required construction is

$$\boxed{(x,y,z)=\bigl(m(m+n),,n(m+n),,-mn\bigr)},$$

where $m,n\in\mathbb Z$, and the number of solutions with $z=1988$ is

$$\boxed{90}.$$

Verification of Key Steps

The equivalence between $(*)$ and $xy+yz+zx=0$ depends only on the identity

$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx).$$

Subtracting $x^2+y^2+z^2$ from both sides leaves

$$2(xy+yz+zx)=0,$$

which is equivalent to $(1)$.

For the parametrization, substitute

$$x=m(m+n),\quad y=n(m+n),\quad z=-mn.$$

A common mistake is to expand incompletely. Factoring $mn(m+n)$ from all three terms gives

$$xy+yz+zx =mn(m+n)\bigl((m+n)-n-m\bigr),$$

and the bracket equals $0$.

For the counting step, equation $(2)$ associates to every integer divisor $d$ the ordered pair

$$\left(d,\frac{1988^2}{d}\right).$$

Different divisors give different pairs. Since $1988^2$ is positive, both positive and negative divisors occur. The positive divisor count is $45$, and including negative divisors doubles the count to $90$. Forgetting the negative divisors would incorrectly produce $45$.

Alternative Approaches

Starting from

$$xy+yz+zx=0,$$

one may solve for $z$:

$$z=-\frac{xy}{x+y}.$$

Choosing integers $u,v,t$ and setting

$$x=t,u(u+v),\qquad y=t,v(u+v),\qquad z=-t,uv$$

produces a more general parametrization. The construction used in the main solution corresponds to $t=1$.

For the second part, one may regard

$$xy+1988x+1988y=0$$

as a rectangular hyperbola and apply Simon's Favorite Factoring Trick directly:

$$(x+1988)(y+1988)=1988^2.$$

The remainder becomes a divisor-counting problem. This approach is essentially equivalent to the main solution but reaches the counting formula immediately.