These are the answers to puzzles that appeared in the Chalkdust newsletters. You can find the puzzles here and sign up for the newsletter here.

### 121

121 is a square number in every base larger than 3.

In base $n$, $$121_n = 1 + 2n + n^2$$ $$=(1+n)^2.$$

### Polya Strikes Out

Source: mscroggs.co.uk
1, 3, 7, 12, 19, …
1, 7, 19, …
1=1; 1+7=8; 1+7+19=27; …
1, 8, 27, …

The final sequence is the cube numbers. To show why, let $n$ be an integer and follow through the process.

Cross out every third number:

1, 2, 3, 4, 5, 6, …, 3n, $3n+1$, $3n+2$, …
1, 2, 4, 5, …, $3n+1$, $3n+2$, …
Find the cumulative sums:
$$1=1$$ $$1+2=1+2=3$$ $$1+2+4=1+2+3+4-3=7$$ $$1+2+4+5=1+2+3+4+5-3=12$$ $$…$$
$$1+2+4+5+…+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$ $$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$ $$=3n^2+3n+1$$ $$1+2+4+5+…+(3n+2)=3n^2+3n+1+(3n+2)$$ $$=3n^2+6n+3$$ $$…$$
1, 3, 7, 12, …, $3n^2+3n+1$, $3n^2+6n+3$, …

Cross out every second number, starting with the second:

1, 3, 7, 12, …, $3n^2+3n+1$, 3n2+6n+3, …

1, 7, …, $3n^2+3n+1$, …

Find the cumulative sums. The $m$th sum is:

$$\sum_{n=0}^{m}3n^2+3n+1$$ $$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$ $$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$ $$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+4m+2)$$ $$=(m+1)(m^2+2m+1)$$ $$=(m+1)(m+1)^2$$ $$=(m+1)^3$$
Hence the numbers obtained are the cube numbers.

These are the answers to puzzles that appeared in the Chalkdust newsletters. You can find the puzzles here and sign up for the newsletter here.

### 121

121 is a square number in every base larger than 3.

In base $n$, $$121_n = 1 + 2n + n^2$$ $$=(1+n)^2.$$

### Polya Strikes Out

Source: mscroggs.co.uk
1, 3, 7, 12, 19, …
1, 7, 19, …
1=1; 1+7=8; 1+7+19=27; …
1, 8, 27, …

The final sequence is the cube numbers. To show why, let $n$ be an integer and follow through the process.

Cross out every third number:

1, 2, 3, 4, 5, 6, …, 3n, $3n+1$, $3n+2$, …
1, 2, 4, 5, …, $3n+1$, $3n+2$, …
Find the cumulative sums:
$$1=1$$ $$1+2=1+2=3$$ $$1+2+4=1+2+3+4-3=7$$ $$1+2+4+5=1+2+3+4+5-3=12$$ $$…$$
$$1+2+4+5+…+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$ $$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$ $$=3n^2+3n+1$$ $$1+2+4+5+…+(3n+2)=3n^2+3n+1+(3n+2)$$ $$=3n^2+6n+3$$ $$…$$
1, 3, 7, 12, …, $3n^2+3n+1$, $3n^2+6n+3$, …

Cross out every second number, starting with the second:

1, 3, 7, 12, …, $3n^2+3n+1$, 3n2+6n+3, …

1, 7, …, $3n^2+3n+1$, …

Find the cumulative sums. The $m$th sum is:

$$\sum_{n=0}^{m}3n^2+3n+1$$ $$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$ $$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$ $$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+4m+2)$$ $$=(m+1)(m^2+2m+1)$$ $$=(m+1)(m+1)^2$$ $$=(m+1)^3$$
Hence the numbers obtained are the cube numbers.

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