## Product Divisible By Sum of Squares

```PROPOSITION:  It's impossible for the product of n distinct primes to
be divisible by the sum of their squares if n < 5.

The proof of this proposition for the cases n = 1, 2, 3, and 4
provides an interesting sequence of progressively more challenging
demonstrations.

CASE n=1:  Since p < p^2, it's self-evident that p^2 cannot divide p.
(Of course, this applies to any integer, not just primes.)

CASE n=2:  If not as self-evident as the case n=1, this case is still
quite trivial, because the identity (p-q)^2 = p^2 + q^2 - 2pq implies
that pq < p^2+q^2.  Thus, pq cannot be divisible by p^2+q^2.  (Again
this applies to any two integers, not just distinct primes.)

CASE n=3:  Here we need to show that pqr is not divisible by p^2+q^2+r^2
for any three distinct primes p,q,r.  If three such primes existed we
would have
pqr = k (p^2 + q^2 + r^2)

for some integer k.  If k is greater than 1 it must be divisible by one
of the three primes on the left, say, r.  But if r is divided out of
both sides, the remaining right side is a multiple of p^2+q^2+r^2, which
we have seen is necessarily greater than pq.  Therefore, k=1.

Now recall that the square of any prime other than 3 is congruent to
+1 (mod 3).  It follows that if none of the primes p,q,r equals 3,
the right side of the above equation is divisible by 3 but the left is
not. On the other hand, if one of the primes is 3, then the left side
is divisible by 3 but the right is not.  Thus the equation cannot be
satisfied in distinct primes (although it can be satisfied by composite
numbers, and also by the set of non-distinct primes p=q=r=3.)

CASE n=4:  For this case we need to prove that the equation

pqrs = k (p^2 + q^2 + r^2 + s^2)

for any integer k cannot be satisfied in distinct primes p,q,r,s.
First, suppose none of the primes is 2.  In that case the left side
is odd but the right side (containing a sum of 4 odd numbers) is even,
so this is impossible.  Let us assign s=2, and note that the sum of
squares must be odd, so k must be even.  Setting k=2m we have

pqr = m (p^2 + q^2 + r^2 + 4)

If m is greater than 1, it must be divisible by one of the primes, say,
r.  Dividing this equation by r we see the right side is a multiple of
p^2+q^2+r^2+4, which is greater than (and therefore cannot divide) pq.
Thus we have m=1.

To complete the proof we will show that the preceding equation with m=1
has no solution in distinct positive integers p,q,r (prime or composite).
Observe first that the smallest of these must be larger than 2, because
otherwise the right hand side is greater than the left.  Also, no two of
p,q,r can be equal, because there is no integer solution of

p^2 = (y^2+4)/(y-2)  =  y + 2 + (8/(y-2))

Now suppose p,q,r is the solution with the least largest value.  We
can arange the integers in order of size, p > q > r > 2, and write
the above equation as a quadratic in p:

p^2 - (qr)p + (q^2 + r^2 + 4)  =  0

It follows that
qr  +-  SR[ (qr)^2 - 4(q^2+r^2+4) ]
p  =  -------------------------------------
2

Since p is real, the quantity in the square root must be positive, and
less than (qr)^2 in magnitude.  This means the square root is less than
qr, so both the roots of this equation are positive integers.  We know
that one of the roots is greater than q, and I will show below that
the other root is less than q, thus giving a solution with a smaller
largest member than p, contradicting the assumption.

The two roots of the quadratic equation, p and p', satisfy the
conditions
p + p' = qr              pp' = q^2 + r^2 + 4

Letting p' denote the smaller of the two roots, the right hand equation

(p')^2  <  pp'  =  q^2 + r^2 + 4  <  3q^2

Thus we have p' < SR(3)q.  Inserting this into the equation p+p'=qr
gives
p + SR(3)q > qr

which implies

p  >  q ( r - SR(3) )

Substituting this into the equation for pp' and dividing both sides
by p gives
q^2 + r^2 + 4
p'  <  -----------------
q ( r - SR(3) )

Evaluating this inequality term by term gives

q           r      r             4
p'  <   ------   +   ---  -------   +  ----------
r-SR(3)       q   r-SR(3)      q(r-SR(3))

Since 2 < r < q, the second term must be less than 2.366.., and the
third term is less than 0.788...  Therefore we have

q
p' <  -------  +  3.154...
r-SR(3)

which shows that p' < q for all q > 15.  Checking the cases with
2 < r < q < 15 shows there are no solutions in this range, so p' must
be less than q.  This implies that p',q,r is another solution in
distinct positive integers with a smaller greatest member than p,

This completes the proof that there are no four (or fewer) primes
whose product is divisible by the sum of their squares.  However,
there do exist many sets of five primes with this property.  The
smallest example is [3,5,11,19,23], because we have

3*5*11*19*23
------------------------------  =  3*23
3^2 + 5^2 + 11^2 + 19^2 + 23^2

For any such set of primes we refer to the ratio of the largest to
the smallest as the "aspect ratio", so the aspect ratio of the above
example is 23/3 = 7.666.   Interestingly, there is a lower limit to
the aspect ratio of such sets.  Most aspect ratios are greater than 6,
but the set [11,13,17,43,61] has an aspect ratio of 61/11 = 5.54, and
the set [11,13,23,43,59] has the even lower aspect ratio of 59/11 =
5.36.  The lowest aspect ratio of all such sets of five primes is for
the set [31,37,47,151,163], with an aspect ratio of 163/31 = 5.258.
In this case we have

31*37*47*151*163
----------------------------------  =  31*37*47
31^2 + 37^2 + 47^2 + 151^2 + 163^2

Coincidentally, the sum of the ten products of three elements of this
set is 2*83*157*163, which is also a multiple of the special quadratic
field prime 163.  This could be regarded as the total volume of the
ten 3-dimensional projections of the 5-dimensional solid with the
minimum possible aspect ratio.
```