Product of 2-digit numbers 
31 people are dancing in a circle formation. Each people’s age is a 2-digit number, and for each digit we know that the units’ digit is equal to the tens’ digit of the person who is in the clockwise position, while the tens’ digit is equal to the units’ digit of the person who is in the counter-clockwise position. Any 2 neighboring digits are different. Check if the product of all 31 ages can be a perfect square.

I don’t know how to start :(
All I can tell is that the digits appear in pairs:
If we have, for example 31 digits, $d_1, d_2, \ldots, d_{31}$, then the ages are:
$$
10 \cdot d_1+d_2,
10 \cdot d_2+d_3,
\ldots,
10 \cdot d_{31}+d_1
$$
and obviously the product is
$(10 \cdot d_1+d_2)
\cdot (10 \cdot d_2+d_3)
\cdots (10 \cdot d_{31}+d_1)$.
But I don’t know how to continue.

D'oh
This is easier. 
Any chain $ab,ba,ab,ba$ of length $4$ will have a perfect square product.
Any chain of $ab, bc, cd, ..... , za$ repeated an even number of times will have a perfect square product.
And there are two digit perfect squares.
The only issue is the even length chains $ab,ba,ab,ba$ end with the same digit they started, and the even number of $ab, bc, cd, ..... , za$ chains must imediately follow each other.
So we need $31 = 4*k + 2j + m$
$31 = 4*7 + 3$
$= 4*4 + 3*4 + 3$
$= (4+1+4+1+ 4+1 + 4)+ 3*4$
So people aged:
$ab,ba,ab,ba$
$ac$
$cd,dc,cd,dc$
$ce$
$ef,fe,ef,fe$
$eg$
$gh,hg,gh,hg$
$hi,jk,ka$
$hi,jk,ka$
$hi,jk,ka$
$hi,jk, ka$
Where $ac,ce,eg$ are perfect squares.
This will do it.
Let $ac = 16; ce=64; eg= 49$ and so long as we can find six able bodied people in their nineties (well, only 4 really) we have it.
Then four people age $6d,d6,6d,d6$.
Then a person aged $64$.
Then four people aged $4f,f4,4f,f4$
Then a person aged $49$.
Then we have four people aged $9g,g9,9g,g9$.  (Two nonagenarians is feasable.  Good for them).
That's 19 people.
The remaining $12$ are $4$ groups of people aged $9h,hi,i1$ (Okay we have six nonagenarians.   A little unlikely but ... heck mor power to them.)
So the product of the ages are $1b^2*b1^2*16*6d^2*d6^2*64*4f^2*f4^2*49*9g^2*g9^2*9h^4*hi^4*i1^4$ is a perfect square.
===== old answer =====
Let $d_1, d_3,..... d_{29} = a$.
Let $d_2, d_4, .... d_{30} = b$.
Let $d_{31} = c$
Ages $d_1d_2,d_3d_4,.....d_{29}d_{30} = 10a + b$
Ages $d_2d_3, d_4d_5,....d_{28}d_{29} = 10b + a$
Age $d_{30}d_{31} = 10b + c$
Age $d_{31}d_1 = 10c + a$.
Product of their ages ard $(10a+b)^{15}(10b+a)^{14} (10b + c)(10c+a)$
We need to find three digits where $(10a + b)(10b+c)(10c + a)$ isa perfect squares.
That shouldn't be hard.
