Talk:Sub-unit squares: Difference between revisions
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:: Therefore, by elimination, the initial perfect square must end in 6.
: If the perfect square ends in 6, then
==An alternative method==
I am going to start by solving a relaxation of the task description. For a z digit number find perfect squares (g*g) ending in 5 such that (g*g)+10<sup>z</sup>/9 is also a perfect square. g will be the sequence of integers between ⌈√10<sup>z</sup>⌉ and ⌊√10<sup>z+1</sup>-1-10<sup>z</sup>/9⌋ ending in 5. For 4 digit numbers the candidates (g) are 25 35 45 55 65 75 85;
For a given g I form the next perfect square by adding 2g+1. So for g=45 I can form a series of perfect squres:
g<sup>2</sup>+91 g<sup>2</sup>+91+93 g<sup>2</sup>+91+93+95 .....
so for 45<sup>2</sup>+1111 to be a sub-unit square 91+93+95+... must equal 1111.
91 184 279 376 475 576 679 784 891 1000 1111 ...
n= 1 2 3 4 5 6
I leave it to the reader to verify that this is not the case for other candidates. In general the value of the nth term (z say) is (2*n - 1)*(2*n + 2*g - 1). Rearranging to find n:
√(g<sup>2</sup>+z) g 1
n = ------ - - + - where z is the value needed i.e. 10<sup>number of digits</sup>/9
2 2 2
In fact it is not necessary to solve for n merely to show that it is an integer. Because I relaxed the tasks' requirement a value of g that passes the above test should be checked for compliance.
--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 13:13, 27 September 2022 (UTC)
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