Rational: $(p^2+1)(q^2+1)(r^2+1)$ is Square

[kaliprasad]

Show that for $p,q,r$ if $pq+qr+rp=1$ then $(p^2+1)(q^2+1)(r^2+1)$ is square of rational number

 

[Euge]

Are $p,\, q$ and $r$ rational numbers?

 

[kaliprasad]

Are $p,\, q$ and $r$ rational numbers?

Thanks for pinpointing. They are rational.

 

[Euge]

Ok then, here's my solution.

Spoiler

We have

$\displaystyle (p^2+1)(q^2+1)(r^2+1)$

$\displaystyle = (pqr)^2 + [(pq)^2 + (qr)^2 + (rp)^2] + (p^2 + q^2 + r^2) + 1$

$\displaystyle = (pqr)^2 + [(pq+qr+rp)^2-2(pqr)(p+q+r)] + [(p+q+r)^2-2(pq+qr+rp)] + 1$

$\displaystyle = (pqr)^2 + [1 - 2(pqr)(p+q+r)] + [(p+q+r)^2-2] + 1$

$\displaystyle = (pqr)^2 - 2(pqr)(p+q+r) + (p+q+r)^2$

$\displaystyle = (pqr - p - q - r)^2$.

Since $p,\, q$ and $r$ are rational, so is $pqr - p - q - r$. Thus we have shown that $(p^2+1)(q^2+1)(r^2+1)$ is the square of the rational number $pqr-p-q-r$.

 

[Opalg]

Another solution:
[sp]Let $x^3 - \lambda x^2 + x - \nu = 0$ be the equation with roots $p,q,r$ (so that $\lambda = p+q+r$ and $\nu = pqr$, the coefficient of $x$ being $qr+rp+pq = 1$).

The equation with roots $p^2,q^2,r^2$ is $x^{3/2} - \lambda x + x^{1 /2} - \nu = 0$, or $x^{1 /2}(x+1) = \lambda x + \nu$, or $x(x+1)^2 = \lambda^2x^2 + 2\lambda\nu x + \nu^2$.

The equation with roots $p^2+1, q^2+1, r^2+1$ is obtained by substituting $x-1$ for $x$ in that last equation, getting $(x-1)x^2 = \lambda^2(x-1)^2 + 2\lambda\nu (x-1) + \nu^2$, or $x^3 - (1+\lambda^2)x^2 + 2\lambda(\lambda-\nu)x - (\lambda^2 - 2\lambda\nu + \nu^2) = 0$. The product of the roots of that equation is $(p^2+1)(q^2+1)(r^2+1) = \lambda^2 - 2\lambda\nu + \nu^2 = (\lambda - \nu)^2$, which is the square of the rational number $\lambda - \nu$.[/sp]

 

[kaliprasad]

2 excellent solutions above. Here is mine and definitely more complex and hence not elegant

Spoiler

$pq+qr+rp = 1$

so $p = \dfrac{1-qr}{q+r}$
Here I may mention that if $q+r = 0$ the we can permute $p,q,r$ such that q+r is not zero

if we chose $q =\ tan\, A$ and $r =\ tan\ B$
we get

$\frac{1}{p} =\dfrac{q+r}{1-qr} =\dfrac {(\tan\ A +\ tan\ B}{(1- \tan\ A\ tan\ B}=\ tan (A+B)$

or $p = \cot (A+B)$

we can chose q and r to be <1 in case we want positive else even -ve also
$(p^2+1) (q^2+1)(r^2+ 1)$
= $\sec^2 A\ \sec ^2 B\ \csc^2 (A+B)$
= $\dfrac{(\sec^2 A\ \ sec^2 B}{\sin^2 (A+B)}$
this is square of reciprocal of $\ sin (A+B)\ cos\ A\ \cos\ B$
$\sin (A+B) \cos\ A\ \cos B$
=$( \sin\ A\ \cos B + \cos\ A\ \sin B)cos\ A\ \ cos\ B$
=$ \sin\ A\ \cos \ A\ \cos ^2B +\ cos^2 A \ \sin B\ \ cos B$
=$\tan\ A\ \ cos ^2 A\ \ cos ^2 B +\ tan\ B \ \
cos ^2 A\ \ cos ^2 B$
= ${(\tan\ A + \tan\ B)}{/(\sec^2 A\ \sec ^2B)}$
= $\dfrac{tan\ A + tan\ B}{(1+ tan ^2 A)(1+ tan ^2B)}$

so $(p^2+1) (q^2+1)(r^2+ 1)$
= $\dfrac{((1+ \tan ^2 A )(1+\tan ^2B)}{(\tan \ A +\ tan B))^2}$

if $ \tan \ A$ and $tan\ B$ that is q and r are rational then

$\dfrac{(1+ \tan ^2 A)(1+\tan ^2B)}{(\tan\ A +\tan\ B)}$ is rational and so $(p^2 +1)(q^2+1)(r^2 +1)$ is the square of a rational no.