algebra challenge problem - Andy's Savings

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Andy's Savings came to one less than twice the square of the difference between the pence and the pounds, all in pence. Given that his savings were between £100 and £200, how much were they?

i.e. let the value be £ abc.de, in pence it would be abcde

$( (abc - de)^2 * 2 ) - 1 = abcde$

I know the answer but I'm not sure how to get it other than trial and error.

The answer is:

£199.99

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For brevity, I'll let the savings (in pence) be $S = a \times 10^4 + b \times 10^3 + c \times 10^2 + d \times 10 + e$

$$(a\times 100 + (b-d) \times 10 + (c-e))^2 = \frac{S+1}{2}$$ $$ \implies (a\times 100 + (b-d) \times 10 + (c-e)) = \sqrt\frac{S+1}{2}$$

Since the left hand side is a 3 digit positive integer, the expression $\frac{S+1}{2} = u^2$ has to be a perfect square of a 3 digit number, with $10000 < S < 20000$. We let $S = 2u^2-1 \implies \frac{10001}2<u^2<\frac{20001}2$. The only perfect square of a 3 digit positive integer falling in this range is $100^2 = 10000$. Thus, $S = 20000-1 = 19999$, which gives us the answer.