Find the 5 digits number that arrange from x,x+1,x+2,3x,x+3 so that the number is a perfect square

Question

I'm having a hard time dealing with this problem and here is my approach:

• The perfect square of 5 digits number must be a 3 digits number ( I put it as a,b,c )
• $abc^2$ get us a 5 digits number that written as $a^2$, $2ab$, $2ac+b^2$, $2bc$, $c^2$

• I see that the 2rd and 4th digit are both even

• If I let $x$ be even or odd the 2rd and 4th digit both won't be the same even

So I conclude that the number is not exist. Is my statement wrong? Please help

Answer 1

If you say the number has five digits -- the first $x$, the second $x+1$, the third $x+2$, the fourth $3x$, the fifth $x+3$ -- then, for $3x$ to be a single non-zero digit, we need $x=1, 2, $ or $3$. Can you take it from here?