We will state and prove two lammas.
LEMMA 1: 10^(2m+1) * n is not a perfect square where n is an integer and n does NOT end in a zero (or, zeros)
PROOF:
If 10^(2m+1) * n is a perfect square, then n must possess the form 10 * t^2., so that n must end in at least one zero.
This is a contradiction.
Consequently, any positive integer ending in an ODD number of consecutive zeroes can NEVER be a perfect square.
LEMMA 2: If the units digit of a perfect square is 6, then the tens digit must be odd.
PROOF:
If a perfect square ends in a 6, its square root must have one of the the forms as: 10t+/-4
Then, the perfect square must be equal to:
(10t+/-4)^2
= 100t^2 +/-80t +16
= 10t(10t+/-8) +16
This is equal to
either:10t((10t+9) + 6
or: 10t(10t-7)+6
In each of these two cases, we observe that the tens digit is ODD.
We can now consider the only other form of the perfect square, that is, 10t+/-6
and arrive at the value of the perfect square as
10t(10t+15) +6 or, 10t(10t-9)
so, the tens digit is ODD in this case as well.
Combining, the foregoing:
The tens digit of the perfect square must always be ODD.
By lemma -1, the perfect square must end in an even number of consecutive zeros.
Then, the nonzero digit preceding the first of these 0s must be a 6.
So, the form of the perfect square must be:
abcd........06000......00
or abcd....6600...00
The tens digit then must be 0 or 6, which are even.
This is a contradiction.
Consequently, there does NOT exist any perfect square having the said structure.
Edited on June 6, 2022, 2:14 am
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