I am looking to find perfect squares $\overline{abcdef}$ with the property that $\overline{ab}$, $\overline{cd}$ and $\overline{ef}$ are perfect squares.
I ran a quick program to find that the only such numbers are $166464=408^2$ and $646416=804^2$, but I can't find a mathematical proof for this, or a way to determine this result without using a computer. Also I am interesed in the phenomenon behind this and if it is pure coincindence that the square roots of the results are palindromes.
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Let $\ x,y,z\ $ run over the two digit perfect squares and check when $$10\ 000x+100y+z$$ is a perfect square as well. In the case of $y$ and $z$, you also have to consider perfect squares $0,1,4,9$ , if $0$ and $1$ count as perfect squares and leading zeros are allowed for $y$ and $z$. Since $x$ is a perfect square , with $\ d:=\sqrt{x}\ $, we can use $$(100d+1)^2=10\ 000 x+200d+1$$ : If $\ 100y+z<200d+1\ $, we cannot have a perfect square , unless we have $\ y=z=0\ $. With this additional restriction, it should be possible to determine the solutions by inspection.
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HINT.- The technical argument you ask can be given by the Square Root Algorithm. You have the square $N^2$ such that $N^2=abcdef$ with $ab,cd,ef$ perfect squares; take $$\sqrt{abcdef}$$ The integers you can take as first digit are $x\in\{4,5,6,7,8,9\}$. Starting with $4$ you have $ab=16$ then you have to calculate z such that $(8z)z$ approximate or equalize $cdef$. Since $89\cdot9=801$ (number with three digits) the second digit is $0$ so you have already $40$ and you should find a last digit to end if an appropriate starting is in fact $4$. Otherwise you choose $5$ and so on (actually you must try with all $x$ above to see the possible answers).
So you have to pick the appropiate last digit in $$801\cdot1\\802\cdot2\\803\cdot3\\804\cdot4\\805\cdot5\\806\cdot6\\807\cdot7\\808\cdot8\\809\cdot9$$ This is a task that can be accomplished at a glance: it is clear that $8$ is the only digit that suits. So you have as first solution $408$ and the first digit chosen was fine.
I believe you that this $408$ and $804$ are the only solutions.
Regarding to the nice topic of palindromes, I think it is due to the presence of the intermediary $0$. Look at this $$(100x+10y+z)^2=10000x^2+100(y^2+20xy+2xz)+z^2+20yz\\(100z+10y+x)^2=10000z^2+100(y^2+20yz+2yz)+x^2+20xy$$ so when $y=0$ $$(100x+z)^2=10000x^2+200(xz)+z^2\\(100z+x)^2=10000z^2+200(xz)+x^2$$ easier and more probably to have palindromes. And, of course $408$ and $804$ are the only ones because you have calculated the only possibilities.
Let $x,y,z\in\{4,5,6,7,8,9\}$ and $\overline{ab}=x^2$, $\overline{cd}=y^2$ and $\overline{ef}=z^2$, and suppose that $$w^2=\overline{abcdef}=10^4x^2+10^2y^2+z^2,$$ for some positive integer $w$. From $4\leq x,y,z\leq9$ we get the (rather loose) inequalities $$(10^2x)^2<10^4x^2+1616\leq10^4x^2+10^2y^2+z^2\leq10^4x^2+8181<(10^2x+11)^2,$$ which show that $$10^2x+1\leq w\leq 10^2x+10.$$ This already narrows the problem down to computing the squares of at most $6\times10=60$ three digit numbers.
Of course $w$ is not divisible by $10$, so $w=10^2x+t$ for some integer $t$ with $1\leq t\leq9$. Then $t^2=\overline{ef}=z^2$ and so $t=z$. Moreover $$2xz=2tx=\overline{cd}=y^2,$$ which shows that $xz$ is twice a square. It follows that $x,z\notin\{5,6,7\}$ or in other words, $x,z\in\{4,8,9\}$. Then either $x=8$ and $z\in\{4,9\}$ or $z=8$ and $x\in\{4,9\}$. Because $$2xz=y^2\leq81,$$ this only leaves $\{x,z\}=\{4,8\}$, corresponding to the solutions you already found.