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
On
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.
On
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 $3334$.
On
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 $\{b, c\}$:
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.
On
This is the same idea as presented in NieDzejkob's post. But the proof was incompletes an contains as far as I can see some errors so I present my complete answer here.
We look for a four digit number such that $${\overline{ab}}^2+{\overline{bc}}^2+{\overline{cd}}^2=\overline{abcd}$$ which means that
$$(10a+b)^2+(10b+c)^2+(10c+d)^2\\=1000a+100b+10c+d, \\ a,b,c,d\in\{0,1,2,\ldots,9\} \tag {1.1}$$ or $$100(a^2+b^2+c^2)+20(ab+bc+cd)+(b^2+c^2+d^2)\\=1000a+100b+10c+d, \\ a,b,c,d\in\{0,1,2,\ldots,9\} \tag{1.2}$$
Maybe one wants to assume that leading digits should not be $0$ but we will not do this.
If $a=n$ then a lower bound $\overline {abcd}$ is $1000n$ and an upper bound is $1000(n+1)$. A lower bound for $(\overline{ab})^2$ is $(10n)^2$ and an upper bound is $(10(n+1))^2$. Note that $$\overline {abcd}-(\overline{ab})^2=(\overline{bc})^2+(\overline{cd})^2$$ So a lower bound for ${\overline{bc}}^2+{\overline{cd}}^2$ is
$$\text{lower}\left({\overline{bc}}^2+{\overline{cd}}^2\right):=\max(1000n-(10(n+1))^2,0)\tag{2.1}$$
and an upper bound is
$$\text{upper}\left({\overline{bc}}^2+{\overline{cd}}^2\right):=\min(1000(n+1)-(10n)^2)\tag{2.2}$$
From this we construct the following table
\begin{array}{|r||r||r|} \tag{Table 1} \hline a&\text{lower}\left({\overline{bc}}^2+{\overline{cd}}^2\right)&\text{upper}\left({\overline{bc}}^2+{\overline{cd}}^2\right)\\ \hline 0&0&1000\\ \hline 1& 600& 1900\\ \hline 2& 1100& 2600\\ \hline 3& 1400& 3100\\ \hline 4& 1500& 3400\\ \hline 5& 1400& 3500\\ \hline 6& 1100& 3400\\ \hline 7& 600& 3100\\ \hline 8& 0& 2600\\ \hline 9& 0& 1900\\ \hline \end{array}
We conclude $$b,c \le 5 \tag 2$$ From $(1.2)$ we get the modulo $2$ equation
$$b^2+c^2+d^2\equiv d \pmod 2$$ and further $$b\equiv c \pmod 2 \tag 3$$
$(2)$ and $(3)$ gives $$ b,c \in \{0,2,4\} \tag{4.1}$$ or $$ b,c \in \{1,3,5\} \tag{4.2}$$
From $(1.2)$ we get also the modulo $5$ equation
$$b^2+c^2\equiv d-d^2 \pmod 5\tag 5$$
The expression $d-d^2 \pmod 5$ can have the following values
\begin{array}{|r||r|}\tag{Table 2} \hline d& d-d^2\mod 5\\ \hline 0 & 0 \\ \hline 1 & 0 \\ \hline 2 & 3 \\ \hline 3 & 4 \\ \hline 4 & 3 \\ \hline \end{array}
we rearrange the table
\begin{array}{|r||r|}\tag{Table 3} \hline b^2+c^2 \pmod 5 & d\\ \hline 0 & 0,1,5,6 \\ \hline 1 & -\\ \hline 2 & - \\ \hline 3 & 2,4,7,9 \\ \hline 4 & 3,8 \\ \hline \end{array}
From this we get
\begin{array}{|r||r|}\tag{Table 4} \hline \{b,c\} & \text{possible}\; d\\ \hline \{0\} & 0,1,5,6\\ \hline \{0,2\} & 3,8 \\ \hline \{0,4\} & - \\ \hline \{2\} & 2,4,7,9 \\ \hline \{2,4\} & 0,1,5,6 \\ \hline \{4\} &- \\ \hline \{1\} &- \\ \hline \{1,3\} & 0,1,5,6 \\ \hline \{1,5\} &- \\ \hline \{3\} &2,4,7,9 \\ \hline \{3,5\} &3,8 \\ \hline \{5\} & 0,1,5,6 \\ \hline \end{array}
The last line of this table can be removed because if $b=5,c=b$ then ${\overline {bc}}^2+{\overline {cd}}^2 \ge 5000$, which contradicts $\text{Table } 1$.
From equation $(1.2)$ we get
$$b^2+c^2+d^2 \equiv 10c+d \pmod {20}$$ and further $$b^2+c^2+10c \equiv -d^2 +d \pmod {20} \tag 6$$ and use this to filter our impossible values from $\text{Table } 4$
We note that
\begin{array}{|r||r|}\tag{Table 5} \hline b^2+c^2 \pmod 5 & d\\ \hline 0 & 0,1,5 \\ \hline 1 & -\\ \hline 2 & - \\ \hline 4 & 3,8 \\ \hline 8 & 4,9 \\ \hline \end{array}
and reduce $\text{Table } 4 $ to
\begin{array}{|r||r|}\tag{Table 6} \hline \{b,c\} &b^2+c^2+10c &\text{possible}\; d\\ \hline \{0\} & 0& 0,1,5\\ \hline \{0,2\} & 4 & 8 \\ \hline \{2\} & 8 & 4,9 \\ \hline \{2,4\} & 0& 0,1,5 \\ \hline \{1,3\}& 0 & 0,1,5 \\ \hline \{3\} & 8 &4,9 \\ \hline \{3,5\} & 4&8 \\ \hline \end{array}
These are $23$ tuples.
From $(1.2)$ we get
$$a=\frac{50-b}{10}\pm\sqrt{\frac{-{{d}^{2}}+\left( 1-20 c\right) d-101 {{c}^{2}}+\left( 10-20 b\right) c-100 {{b}^{2}}+2500}{100}}\tag{6}$$
The discriminant (this is expression under the square root) must be a perfect square. The calculation of the discriminant seems to be a little bit cumbersome without the usage of a calculator. But if a number is a perfect square it has a square root modulo every module $m$. So we reduce this table by checking if the discriminant has a square root modulo $9$ and if this is the case we check further it it has a square root modulo $11$. The calculation with these moduli is simple because we have to calculate only with one digit numbers except for $10 \pmod{11}$.
The discriminant modulo $9$ is
$$( 1-2 c-d) d+( 1-2 b-2c) c-b^2-2$$
Modulo $11$ it is
$$( 1+2 c -d) d+( 2 b-1-2c) c-b^2+3$$
From this we create $\text{Table 7}$. $r_9$ is the discriminant modulo $9$. I the root exists, this is if $r_9 \in \{0,1,4,7\}$ then we also calculate $r_{11}$, the discriminant modulo $11$. If the root modulo $11$ exists we finally solve the quadratic equation modulo $9$ and $11$. We have
$$a=5-b\pm \sqrt{( 1-2 c-d) d+( 1-2 b-2c) c-b^2-2}\pmod 9$$
and
$$a=5+b\pm \sqrt{( 1+2 c -d) d+( 2 b-1-2c) c-b^2+3} \pmod {11}$$
We add $9$ to the possible solutions modulo $9$ if if contains $0$ because $0\equiv 9 \pmod 9$. This is the column $a_9$. We remove $10$ from the possible solutions modulo $11$, because $10$ is not a digit. This is the column $a_{11}$.
We get the following table the following table:
\begin{array}{|r||r|}\tag{Table 7} \hline b&c&d&r_9&r_{11}&a_9&a_{11} \\ \hline 0 & 0 & 0 & 7 & 3 & 0, 1, 9 & 0 \\ \hline 0 & 0 & 1 & 7 & 3 & 0, 1, 9 & 0 \\ \hline 0 & 0 & 5 & 5 \\ \hline 0 & 2 & 8 & 3 \\ \hline 1 & 3 & 0 & 3 \\ \hline 1 & 3 & 1 & 6 \\ \hline 1 & 3 & 5 & 7 & 8 \\ \hline 2 & 0 & 8 & 1 & 9 & 2, 4 & 4 \\ \hline 2 & 2 & 4 & 6 \\ \hline 2 & 2 & 9 & 7 & 5 & 7, 8 & 0, 3 \\ \hline 2 & 4 & 0 & 4 & 1 & 1, 5 & 6, 8 \\ \hline 2 & 4 & 1 & 5 \\ \hline 2 & 4 & 5 & 7 & 10 \\ \hline 3 & 1 & 0 & 0 & 8 \\ \hline 3 & 1 & 1 & 7 & 10 \\ \hline 3 & 1 & 5 & 6 \\ \hline 3 & 3 & 4 & 1 & 3 & 1, 3 & 2, 3 \\ \hline 3 & 3 & 9 & 1 & 6 \\ \hline 4 & 2 & 0 & 5 \\ \hline 4 & 2 & 1 & 1 & 8 \\ \hline 4 & 2 & 5 & 1 & 4 & 0, 2, 9 & 0, 7 \\ \hline \end{array}
The Chinese Remainder Theorem tells us that for each tuple $b,c,d$ each element of $a_9$ can be combined with an element of $a_{11}$ and that gives us an element $a$ of $\mathbb {Z_{99}}$ that is a solution of the equation $\pmod{99}$. But we are only interested in value $a$ that are in $0,1,2,\ldots,9$. This is only the case if the number is in $a_9$ and $a_{11}$.
\begin{array}{|r||r|}\tag{Table 8} \hline b&c&d&a \\ \hline 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 1 & 0 \\ \hline 2 & 0 & 8 & 4 \\ \hline 3 & 3 & 4 & 3 \\ \hline 4 & 2 & 5 & 0 \\ \hline \end{array}
But it is still not guaranteed that such a tuple is a solution because the discriminant may not be a perfect square and therefore the root may not exist in $\mathbb Z$. So we have to check $(1.1)$ for each of these possible solutions and find
$$0000=00^2+00^2+00^2$$ $$0001=00^2+00^2+01^2$$ and finally $$3334=33^2+33^2+34^2$$
Faster and shorter in Python:
for n in range(10000):
s=('0000'+str(n))[-4:]
if int(s[:2])**2+int(s[1:3])**2+int(s[2:])**2==n:
print(n)
Output
0
1
3334
It's not a pretty answer, but here is a brute-force solution in Python