1) Clearly y >=1 because the LHS can never be negative. X can be any integer.
2) The RHS must always have a zero for the units digit, therfore is divisible by 10.
3) Given 2), x^2 must have either a 0, 4, or 5 in it's units digit.
4) If x^2 ends in 0, x must be of the form 10^i, i=integer>=1
5) If x^2 ends in 5, x must be in the form 5^j, j=integer>=1
6) If x^2 ends in 4 - things are more complex. At a minimum, x can be of the form 2^(2n-1) or 8^(2n-1), where n=is a positive integer. Upon inspection, the solutions using powers of 8 are a subset of those that are a powers of 2.
My head is starting to spin.
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Posted by Kenny M
on 2021-01-03 07:47:10 |
Some Thoughts
