Adding 1 to the product of four consecutive positive integers always results in a perfect square. The first 2017 such square numbers can be found:

Question

1 x 2 x 3 x 4 + 1=25=5^2
2 x 3 x 4 x 5 + 1=121=11^2
3 x 4 x 5 x 6 + 1=361 = 19^2
2017 x 2018 x 2019 x 2020 + 1=16600254584281=4074341^2
In this list of 2017 numbers: 5, 11, 19, ........, 4074341
Whose squares are found in this way, how many have last digit equal to 1?

Answer 1

The pattern itself is easy to prove
Let the 4 consecutive integers be n-1, n, n+1, and n+2
Did you notice the pattern?
Then prove
(n-1)(n)(n+1)(n+2) = (n(n+1) - 1)^2
which is easy

List the first 15 or so cases , in the first 10 cases , the only ones where the square ends in a 1 are .,.
2*3*4*5+1= 11^2
5*6*7*8+1 = 41^2
7*8*9*10+1 = 71^2
10*11*12*13+1 = 131^2
In each group of 10, this happens 4 times
From 1 to 2010 this happens 201*4 or 404 times
From 2011 to 2017 we have 3 more
For a total of 407 such cases

Check my arithmetic

Answer 2

Noticed my own aritmetic error
201*4= 804, not 404

So the total is 807