Find a four digit number which is a perfect square. The last two digits are equal and the first two digits are equal of the number.

Question

asked Mar 20, 2017 200k views

1 Answer

Muhammed Ozdogan · Answer 1 · 2017-03-24T05:43:48+0000

$1000x+100x+10y+y=k^2\qquad (k\in \mathbb{Z})\\ 1100x+11y=k^2\\ 11(100x+y)=k^2$

11 is a prime number, so for
k^2

to be a perfect square,
100x+y

must be also equal to 11. But
11^2=121

which is not even a four digit number. Therefore there is no such number.