FMM10

Friday Morning Math Problem

Square Free Sequence

Prove that no number in the sequence

11, 111, 1111, 11111, ...

is the square of an integer.

Solution

If s is a number in the sequence, s must have the form $11+100m$ where m is a non-negative integer. This can be rearranged into two parts: the part divisible by 4, and the part not divisible by 4: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 11 + 100m = 100m + 8 + 3 = 4 (25m + 2) + 3 } which means that when divided by 4, all numbers in the sequence have a remainder of 3. Furthermore, we know that all squares are of the form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 4n^2 } or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 4n^2 + 4n + 1 } and therefore leave remainders of 0 or 1 when divided by 4. Thus, a number s in the sequence that always has a remainder of 3 cannot be a square.