Finding a four-digit number $\overline{abcd}$ such that $(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $ Consider a four digit number $\overline{abcd}$ such that $$(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $$
$\overline{ab}$, $\overline{bc}$ and $\overline{cd}$ are two digit numbers
What is this number ?
What I have tried is
$$(10a+b)^2+(10b+c)^2+(10c+d)^2=\overline{abcd} $$ then I used $(a+b)^2$ formula but this method didn't help.
I need mathematical solution
 A: 
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did a quick search where I look for values that can be written as $1000\text{a}+100\text{b}+10\text{c}+\text{d}$ where $\left\{\text{a},...,\text{d}\right\}\in\left\{1,...,9\right\}$

I wrote and ran some Mathematica-code:
In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1000*a + 100*b + 10*c + 
      1*d == (a*b)^2 + (b*c)^2 + (c*d)^2], 
   1000*a + 100*b + 10*c + 1*d, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1,
    9}, {d, 1, 9}] //. {} -> Nothing

Running the code gives:
Out[1]={{{{1284}}}}

Because:
$$\left(1\cdot2\right)^2+\left(2\cdot8\right)^2+\left(8\cdot4\right)^2=1284\tag1$$
That is the only four digit number with that property.

If you mean that two digit number thing, I got:
In[2]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1000*a + 100*b + 10*c + 
      1*d == (a*10 + b)^2 + (b*10 + c)^2 + (c*10 + d)^2], 
   1000*a + 100*b + 10*c + 1*d, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1,
    9}, {d, 1, 9}] //. {} -> Nothing

Out[2]={{{{3334}}}}

Because:
$$\left(10\cdot3+3\right)^2+\left(10\cdot3+3\right)^2+\left(10\cdot3+4\right)^2=3334\tag2$$
That is the only four digit number with that property.
A: Denote $S = (\overline{bc})^2+(\overline{cd})^2$. Note that $S < 3500$ — otherwise, we have $a \geq 3$, therefore $\overline{abcd} \geq 3500 + 900 = 4300$, so $a \geq 4$, and if we continue we reach a contradiction.
In particular, this implies $b, c < 6$.
Taking our original equation mod 10, we obtain
$$b^2 + c^2 + d^2 \equiv d \pmod{10}$$
$$b^2 + c^2 \equiv - d(d-1) \pmod{10}$$
Checking each value of $d$ separately, we see that the right side can only have values 0, 4, and 8. Similarily, considering the quadratic residues mod 10 (all of which happen to be reachable with $b, c < 6$), you'll find five possible pairs for $\{b, c\}$:

*

*3, 1

*3, 3

*3, 5 (which can be eliminated with the $S < 3400$ condition)

*2, 0 (which violates the condition that the numbers being squared should have two digits)

*2, 4

*2, 2

Each of those also has a small subset of possible values of $d$.
While this gets us near the threshold where checking the cases by hand becomes possible, I personally wouldn't want to do that.
I'd try finding a lower bound for $S$, to complement or upper bound of 3400 and help prune some of these possibilities.
A: It's not a pretty answer, but here is a brute-force solution in Python
>>> from itertools import product
>>> digits = '0123456789'
>>> for a,b,c,d in product(digits, digits, digits, digits):
...     if int(a+b)**2 + int(b+c)**2 + int(c+d)**2 == int(a+b+c+d):
...         print(a + b + c + d)
... 
0000
0001
3334

A: It could be done in a simpler way using  Mathematica
FindInstance[(10 a+b)^2+(10 b+c)^2+(10 c+d)^2==1000 a+100 b+10c+d
 &&0<a<10&&0<b<10&&0<c<10&&0<d<10,{a,b,c,d},Integers,2]

and get
$$\{\{a\to 3,b\to 3,c\to 3,d\to 4\}\}$$
the number is $3334$.
