68062500 = 8250^2
with p = 367, q =433.
I can prove
that 34811956 is the smallest square that gives a solution
to the problem.
depends on the follwing :
Theorem : if s
is a square with k digits and if p is a prime such that
the concatenation p&s is a square,
then p&s is
the square of a number less than 10^k.
Assume s has
k digits and s = n^2.
Let p be a
prime such that p&s is a square. Then we can write :
p&s = p*10^k +
s = (x*10^k + t)^2
where t is a
number < 10^k (and >= n) whose square's last k digits
coincide with s.
I.e. we can
write t^2 as s + m 10^k.
Then : p*10^k
+ s = (x^2)*(10^(2k) + 2 t x 10^k + t^2 = (x^2)*(10^(2k) + 2
t x 10^k +m 10^k + s
after some trivial simplification :
(x^2)*(10^k) + 2 t x + m.
Thus, p can be
expressed as the value of some quadratic polynomial in x.
the discriminant of that polynomial is a perfect square !
polynomial has rational roots and thus can be written as
(a x + b)(c x + d) with a, b, c, d positive numbers (Gauss's
Lemma on Polynomials).
It is clear
that this equals a prime if and only if x = 0 and b or d
= 1. But, this is precisely Q.E.D.
This means that
we may find p as follows : list all elements t between
n and 10^k whose square ends in s.
Then, look at
the digits of t^2 that precede s : these should form p..
Of course, once
we found such a p, we still have to see wether or not s&p
when s = 36, we have only four numbers less than 100 that
have a square ending in 36 :
6, 44, 56 and
94 with squares resp : 36, 1936, 3136 and 8836. But only
3136 is a 'good'.
In our search
for solutions, we also can restrict ourselves to squares
that are divisible by 9. This can be deduced easily if we
look at the equations
p*10^k + s =
square and s*10^t + p = prime (mod 3).
Using our PC,
it is then easy to see that 3481956 is the smallest square.