128 is the largest number which is not the sum of distinct squares define?
Answers
Step-by-step explanation:
no its not.................
Step-by-step explanation:
First, you show that you can express all numbers 129 though something convenient as a sum of distinct squares just by making a list. I will choose 4900 and assume that we have a list that runs that high. We do the proof by strong induction. Assume that all numbers in the range [129,k] can be expressed as a sum of distinct squares, where k≥4900. We shall show that k+1 can be so expressed. Let m=⌊k+1−−−−√⌋−1. We can express k+1=m2+p, where p≥129, so it can be expressed as a sum of distinct squares. Since p<m2, there is no conflict with the new square, and k+1 can be expressed as a sum of distinct squares. You could lower the bound of the specific list below 3600 if you think harder about what square to subtract. I chose 3600 because the spacing of squares attains 129 at that point.
Added: the list you make can run only through [129,297]. We can be lazier in the first part with more work in the second. 297 is a risk because 297=169+128, so the largest square you can subtract and stay larger than 129 is 144 and the expression for 153 could involve 144. In fact 144+9=153, but you can also do 100+49+4. Starting with 298 you can always subtract a square that leaves at least 129 and is greater than half the starting number, so the strong induction goes through.