Kvant Math Problem 1088
The condition is
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m41s
Source on kvant.digital
Problem
Prove that if the numbers $p$, $q$, $r$ are rational and $pq+qr+pr=1$, then $(1+p^2)(1+q^2)(1+r^2)$ is the square of a rational number.
István Varga (Romania)
Exploration
The condition is
$$pq+qr+rp=1.$$
We must show that
$$(1+p^2)(1+q^2)(1+r^2)$$
is a rational square.
A direct expansion gives
$$(1+p^2)(1+q^2)(1+r^2) =1+(p^2+q^2+r^2)+(p^2q^2+q^2r^2+r^2p^2)+p^2q^2r^2.$$
Since
$$p^2q^2+q^2r^2+r^2p^2=(pq+qr+rp)^2-2pqr(p+q+r),$$
the given condition replaces this term by
$$1-2pqr(p+q+r).$$
Also
$$p^2+q^2+r^2=(p+q+r)^2-2.$$
Substituting,
$$(1+p^2)(1+q^2)(1+r^2) =(p+q+r)^2-2pqr(p+q+r)+p^2q^2r^2.$$
This factors as
$$(p+q+r-pqr)^2.$$
Before trusting this identity, test a few examples. If $p=q=r$, then $3p^2=1$, which gives irrational $p$, so rational examples must be different. Take $p=1$, $q=0$, $r=1$. Then $pq+qr+rp=1$. The product equals
$$2\cdot1\cdot2=4,$$
and
$$p+q+r-pqr=1+0+1-0=2.$$
The square is indeed $4$.
Take $p=2$, $q=\frac13$. Then
$$\frac23+r!\left(2+\frac13\right)=1,$$
hence $r=\frac17$. The product is
$$5\cdot\frac{10}{9}\cdot\frac{50}{49} =\frac{2500}{441},$$
while
$$p+q+r-pqr =2+\frac13+\frac17-\frac{2}{21} =\frac{50}{21},$$
whose square is also $\frac{2500}{441}$.
The only potentially dangerous step is the algebraic factorization. Once that identity is established, rationality is immediate because $p+q+r-pqr$ is rational.
Problem Understanding
We are given rational numbers $p,q,r$ satisfying
$$pq+qr+rp=1.$$
The task is to prove that the number
$$(1+p^2)(1+q^2)(1+r^2)$$
is the square of a rational number.
This is a Type B problem. A statement is given and must be proved.
The core difficulty is finding a factorization of the product that uses the condition $pq+qr+rp=1$. After the correct algebraic identity is found, the conclusion follows immediately from the rationality of $p,q,r$.
Proof Architecture
The first lemma is that
$$(1+p^2)(1+q^2)(1+r^2) =1+(p^2+q^2+r^2)+(p^2q^2+q^2r^2+r^2p^2)+p^2q^2r^2,$$
which follows from direct expansion.
The second lemma is that
$$p^2q^2+q^2r^2+r^2p^2 =(pq+qr+rp)^2-2pqr(p+q+r),$$
obtained by expanding the square.
The third lemma is that
$$p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+rp),$$
which is the standard identity for the square of a sum.
The fourth lemma is that, after substituting $pq+qr+rp=1$ into the preceding formulas,
$$(1+p^2)(1+q^2)(1+r^2) =(p+q+r-pqr)^2.$$
The hardest point is verifying the algebra leading to the final factorization, because a sign error in the mixed term would invalidate the conclusion.
Solution
Expanding the product gives
$$(1+p^2)(1+q^2)(1+r^2) =1+(p^2+q^2+r^2)+(p^2q^2+q^2r^2+r^2p^2)+p^2q^2r^2.$$
Next,
$$(pq+qr+rp)^2 =p^2q^2+q^2r^2+r^2p^2+2pqr(p+q+r),$$
hence
$$p^2q^2+q^2r^2+r^2p^2 =(pq+qr+rp)^2-2pqr(p+q+r).$$
Also,
$$(p+q+r)^2 =p^2+q^2+r^2+2(pq+qr+rp),$$
so
$$p^2+q^2+r^2 =(p+q+r)^2-2(pq+qr+rp).$$
Using the condition $pq+qr+rp=1$, we obtain
$$p^2+q^2+r^2=(p+q+r)^2-2$$
and
$$p^2q^2+q^2r^2+r^2p^2 =1-2pqr(p+q+r).$$
Substituting these expressions into the expanded product yields
$$\begin{aligned} (1+p^2)(1+q^2)(1+r^2) &=1+\bigl((p+q+r)^2-2\bigr) +\bigl(1-2pqr(p+q+r)\bigr) +p^2q^2r^2\ &=(p+q+r)^2-2pqr(p+q+r)+p^2q^2r^2. \end{aligned}$$
The right-hand side is
$$(p+q+r-pqr)^2.$$
Therefore
$$(1+p^2)(1+q^2)(1+r^2) =(p+q+r-pqr)^2.$$
Since $p,q,r$ are rational, the number $p+q+r-pqr$ is rational. Hence
$$(1+p^2)(1+q^2)(1+r^2)$$
is the square of a rational number.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the identity
$$p^2q^2+q^2r^2+r^2p^2 =(pq+qr+rp)^2-2pqr(p+q+r).$$
Expanding the square from scratch,
$$(pq+qr+rp)^2 =p^2q^2+q^2r^2+r^2p^2 +2pq\cdot qr+2qr\cdot rp+2rp\cdot pq.$$
The three cross terms are
$$2pq^2r+2pqr^2+2p^2qr =2pqr(p+q+r),$$
which gives the formula exactly.
The second delicate step is the final factorization. Starting from
$$(p+q+r)^2-2pqr(p+q+r)+p^2q^2r^2,$$
set
$$A=p+q+r,\qquad B=pqr.$$
Then the expression becomes
$$A^2-2AB+B^2=(A-B)^2,$$
hence
$$(p+q+r-pqr)^2.$$
A common mistake is to substitute
$$p^2+q^2+r^2=(p+q+r)^2-2$$
without first using the condition $pq+qr+rp=1$. The correct general identity contains the factor $2(pq+qr+rp)$, and omitting it prematurely changes the computation.
Alternative Approaches
A symmetric-polynomial approach introduces
$$s=p+q+r,\qquad t=pqr.$$
Since $pq+qr+rp=1$, the elementary symmetric polynomials are $s$, $1$, and $t$. Expanding
$$(1+p^2)(1+q^2)(1+r^2)$$
and expressing everything through the elementary symmetric polynomials gives
$$1+(s^2-2)+(1-2st)+t^2=(s-t)^2.$$
This reaches the same factorization in a more systematic way.
Another route starts from the identity
$$(1+p^2)(1+q^2)(1+r^2)-(p+q+r-pqr)^2 =(pq+qr+rp-1)^2.$$
Expanding both sides verifies the identity directly. Under the condition $pq+qr+rp=1$, the right-hand side vanishes, yielding the desired result immediately. The main solution is preferable because it derives the factorization using the given condition at an earlier stage and requires less algebraic expansion.