Is a,b are non zero digits, then prove that a number of the form a00...00b can never be a perfect square

Dear student
Here is your answer
In this case,

the place value of “a” should be of the form of [(10^n)^(2)]a.

So ,

If we expand it then,

[(10^n)^(2)]a + b = (c)^(2)

or, [(10^n)^(2)] = [(c)^(2) - b]/a

here “a” & “b” are digits.

For the above equality to hold,

the only digit for “a” which holds equality is 1.

Therefore the eqn becomes

[(10^n)^(2)] = [(c)^(2) - b]

or, [(10^(n)^(2)] + b =(c)^(2)

Now since “a” & “b” are non zero & there are no digits which satisfies the equality so (c)^(2) isn’t possible.

note: (10)^(2) = (26)^(2) - (24)^(2)

26 & 24 are not digits.


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