# Sunday Times Teaser 2834 – Degrees of Freedom

### by Stephen Hogg

#### Published: 15 January 2017 (link)

I bought an odd thermometer from an old curiosity shop. On its linear scale the freezing point and boiling point of water were higher than they are on the centigrade scale. In fact the freezing point was a prime number and, higher up the scale, the boiling point was a perfect square. There was only one number on the scale where it actually agreed with the corresponding centigrade temperature. That number was the negative of an odd prime (and not the same prime as the one mentioned earlier).

On this new scale, what are the freezing and boiling points of water?

Let $$t_f$$ (an odd prime) and $$t_b$$ (a perfect square) be the freezing and boiling points of water on the odd thermometer’s scale. To convert its reading ($$t$$) into centigrade we have to use: $100(t – t_f)/(t_b-t_f)$ So the point at which the two scales give an equal value is given by:$t=100t_f/(100+t_f-t_b)$ Let this be the negative of an odd prime $$p$$, which leads to the equation: $t_b=100+t_f+100t_f/p$ Since $$t_f$$ is an odd prime and $$t_b$$ is an integer perfect square (say $$n^2$$), $$p$$ must be 5 (it is greater than 2 and not equal to $$t_f$$), which gives:$t_f=(n-10)(n+10)/21$ Since $$t_f$$ is an odd prime, $$(n-10)(n+10)$$ must be divisible by 21. But if one of these terms was divisible by 3 and the other by 7, then $$t_f$$ would be the product of two integers and hence not a prime (if one of these integers is 1, it can be shown that the other cannot be prime) . So one of these terms must be divisible by 21.
If $$(n+10)$$ is divisible by 21, then $$t_f=k(21k-20)$$ for some integer $$k$$, which cannot be prime. So $$(n-10)$$ must be divisible by 21, giving $$t_f=41$$ and $$t_b=961$$.