^{2}+ 4

^{2}= 5

^{2}, in which all three numbers are consecutive, and another similar coincidental-seeming equation 10

^{2}+ 11

^{2}+ 12

^{2}= 13

^{2}+ 14

^{2}. These are the first two elements of a sequence of similar equations, with k+1 squares on the left-hand side and k squares on the right-hand side, which she presents there.

Still perhaps coincidental is the fact that 3

^{2}+ 4

^{2}= 5

^{2}and 3

^{3}+ 4

^{3}+ 5

^{3}= 6

^{3}. Of course once you see these it's tempting to check if 3

^{4}+ 4

^{4}+ 5

^{4}+ 6

^{4}= 7

^{4}. (It doesn't.) In fact, for n ≥ 4,

3

^{n}+ 4

^{n}+ ... + (n+2)

^{n}< (n+3)

^{n}

and here's a proof. I'll prove it for n ≥ 5. It suffices to show that

(3/(n+3))

^{n}+ (4/(n+3))

^{n}+ ... + ((n+2)/(n+3))

^{n}< 1.

Call the left-hand side f(n). But the last term on the right-hand side is decreasing in

*n*, and for n ≥ 5, is at most 16807/32768 (its value at 5). Now, we can write the left-hand side of the above inequality as

((n+2)/(n+3))

^{n}(1 + ((n+1)/(n+2))

^{n}+ (n/(n+2))

^{n}+ ... + (3/(n+2))

^{n})

and each term in the second factor is at most ((n+1)/(n+2))

^{n}times the previous one. So we have

(1 + ((n+1)/(n+2))

^{n}+ (n/(n+2))

^{n}+ ... + (3/(n+2))

^{n}) > (1 + r + r

^{2}+ r

^{3}+ ...)

where r = ((n+1)/(n+2))

^{n}. Of course the right-hand side of the previous inequality is 1/(1-r). r is positive and decreasing in

*n*, so 1/(1-r) is as well. Since n ≥ 5, 1/(1-r) is bounded above by its value at 5, which is 16807/9031. Thus for n ≥ 5,

f(n) < (16807/32768)(16807/9031)

and the right-hand side is easily checked to be less than 1.

For n = 4 this proof doesn't work, because I threw away a bit too much in bounding the sum of a finite series by the sum of an infinite one, but it's easy to check.

In fact, f(n) approaches 1/(e-1) as n goes to infinity.

**Edit (8:41 pm):**: See also The Everything Seminar.

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