The sequence is the squares of numbers which leaves a remainder 56 when divided by 100 that is
$n \equiv 56 \pmod {100}$
We can look at the numbers mod 100 and then find the numbers from table lookup. However this is not interesting
To keep the things simple let us find the numbers mod 10 and then make it to numbers mod 100.
Let us look at the number mod 10 and find the square mod 10 we get (1st is number mod 10, 2nd is number square mod 10 $(0,0),(1,1),(2,4),(3,9),(4,6),(5,5),(6,6),(7,9),(8, 4),(9,1)$
So then umber is 4 mod 10 or 6 mod 10
That is number is of the form 10n+ 4 or 10n+ 6
Now we need to find n in both the forms.
We can take the x as 2 digit number . adding 100 to x shall not change the 2 digits mod 100
This is because
$(100n + k)^2 = 10000n^2 + 200nk + k^2$
so $(100n + k)^2 = k^2\pmod {100}$
Consider first form $x = 10n + 4$
Now $(10n+4)^2 = 100n^2 + 80n + 16$
so $80n + 16 = 56\pmod {100}$
or $80n = 40 \mod 100$
dividing by 20 on both sides
$2n = 1\pmod 5$
or $n = 3 \mod 5$ or $n \in \{3,8\}$ as we are looking for single digit number
So the number is $34 \pmod {100}$ or $84 \pmod {100}$
we can combine both to get $x \equiv 34\pmod {50}$
Considerfnext form $x = 10n + 6$
Now $(10n+6)^2 = 100n^2 + 120n + 36$
As remainder is 56
so $20n + 16 = 56\pmod {100}$
or $20n = 20 \mod 100$
dividing by 20 on moth sided
$n = 1\pmod 5$
or $n = 1 \mod 5$ or $n \in \{1,6\}$ as we are looking for single digit number
So the number is $16 \pmod {100}$ or $66 \pmod {100}$
So we have $x \equiv 16 \pmod {50}$ or $x \equiv 34 \pmod {50}$
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