Using only calculations you can do in your head, can you show that the century starting in 3500 is the first that will not contain a year that is a perfect square?

For the algebra, it may be simpler to use the fact that (x+1)^2 – x^2 = 2x+1

Nope.

3600 = 60 squared. 59 squared = 60 squared – 60–59 = 60 squared minus 119.

The difference between the square of a number and its next higher square is actually twice that number +1.

60^2 -> 3600
59^2 = 3600 – 119 = 3481
58^2 = 3491 – 117 = 3364
57^2 = 3364 – 115 = 3249
56^2 = 3249 – 113 = 3136
55^2 = 3136 – 111 = 3025
54^2 = 3025 – 109 = 2916
53^2 = 2916 – 107 = 2809
52^2 = 2809 – 105 = 2704
51^2 = 2704 – 103 = 2601
50^2 = 2601 – 101 = 2500
49^2 = 2500 – 99 = 2401

note that from there on down the difference between the current and the next lower square is less than 100. So it will be impossible to skip a century with going down one square.

Since every century between 2500 and 3500 has a s square int them and the period from 3500 through 3599 doesn’t, this is the first period to not have square in them.

@whitenoise , You got most of what I was looking for. You can also avoid computing the squares of the numbers greater than 2500, which I will show later.

This solved your question without having to think much about it… call it laziness. :-)

I look forward to your explanation.

A century actually starts on xx01, so the first century to not have a perfect square would be the 2500’s.

60*60=3600
59*59=3481

@amujinx correct,
50*50=2500
51*51=2601

What’s it like to be good with math AND big words?

Guys…playing authistic?

You knew perfectly well what the question was meant to ask…

I would mentally compute squares as

x^2 = (x-y)(x+y) + y^2

The value of y can be picked strategically. I pick y = x – 50.

61^2 = 50*72 + 121 = 3721
60^2 = 50*70 + 100 = 3600
59^2 = 50*68 + 81 = 3481
58^2 = 50*66 + 64 = 3364

Here, you can see that a century is skipped whenever the hundreds digit decreases by at least two. The first term of the formula (the 50 times N) always decreases in value by one hundred. So, a century will be skipped whenever the second term (the small square) decreases its hundreds digit.

Below 59^2, this is not possible, as the second term is already less than 100. So, the jump between 3481 and 3600 is the smallest place where an entire century is skipped.

The 36th Century would begin with 3501, so 3600 would qualify, but since you phrase the question this way, the easy trick is:

60×60 = 3600.

Subtract 60 + 59 from 3600, and you have a number under 3500, 3481 to be exact.

Yes, I studied accounting in school

@filmfann , Precisely what I had in mind for the third step.

The second step is just a matter of counting First realize that starting with 2500, every century has at most one perfect square, Stated differently, every century has 1 or 0 perfect squares, 2500 to 3599 is 11 centuries. The perfect squares in this period are the 10 squares of the numbers from 50 to 59. That means 10 of the centuries have 1 perfect square and 1 century does not have any. Then using @filmfann ‘s approach, you can show that the last century in the group is the one missing the perfect square.