### Euler's Squares

Euler found four whole numbers such that the sum of any two of the numbers is a perfect square. Three of the numbers that he found are a = 18530, b=65570, c=45986. Find the fourth number, x. You could do this by trial and error, and a spreadsheet would be a good tool for such work. Write down a+x = P^2, b+x = Q^2, c+x = R^2, and then focus on Q^2-R^2=b-c which is known. Moreover you know that Q > sqrtb and R > sqrtc . Use this to show that Q-R is less than or equal to 41 . Use a spreadsheet to calculate values of Q+R , Q and x for values of Q-R from 1 to 41 , and hence to find the value of x for which a+x is a perfect square.

### Odd Differences

The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.

### Substitution Cipher

Find the frequency distribution for ordinary English, and use it to help you crack the code.

##### Stage: 4 Challenge Level:

We have yet to receive a solution that uses only Level 4 mathematics, (so a solution that doesn't use logs) and so this problem has become Toughnut! See if you can crack it and send us your solution!

We have now received some excellent solutions for this Toughnut question. Patrick from Woodbridge School solved the first question by making a list of each years values:

Let x be the original value.
To add 10%, we multiply by 1.1, and we are trying to reach 2x. Thus, we get
x
1.1x
1.21x
1.331x
1.4641x
1.61051x
1.771561x
1.9487171x
2.14358881x
So, after 8 years, the sum has doubled.

To reduce a value by 10%, we multiply by 0.9, and we must reach 0.5x. Thus, we get
x
0.9x
0.81x 11
0.729x
0.6561x
0.59049x
0.531441x
0.4782969x
so after 7 years the sum has halved in value.

Alexandra, Shannon, Gemma, Katie, Ruby and Caitlin from Herts and Essex High School for Girls also worked out the problem in this way correctly. Well done!

Iain from St. Gregory's Catholic High and Daniel from King's College, Madrid used logarithms to solve the problem:

If a sum increases by 10% each year this means that the value of the previous year is multiplied by 1.1 (it is 110% of what it was before). Therefore for an initial value "a" the progression will be:
Year:        0                   1                         2                                 3                             ...
$a$             $a\times1.1$        $(a\times1.1)\times1.1$       $((a\times1.1)\times1.1)\times1.1)$
= $a$             $a\times(1.1)$          $a\times(1.1^2)$                   $a\times(1.1^3)$                       ...

A sum invested gains 10% each year:
$1.1^n = 2$                                                             (1)
$n log 1.1 = log2$
$n = \frac{log2}{log1.1} = 7.273$                                        (4.s.f.)
It will therefore be 8 years before the sum is effectively doubled.

An object depreciates in value by 10% each year:
$0.9^m = 0.5$                                                         (2)
$mlog0.9 = log0.5$
$m = \frac{log0.5}{log0.9} = 6.579$                                      (4.s.f.)
It will therefore be 7 years before the sum is effectively halved.

Patrick from Woodbridge School went on to explain why it takes different time for the value to double and half:

Why aren't these two answers the same?
The answers are not the same as, when 10% is added to a value, less than 10% must be taken off as the 10% of the new number is larger. This can be summed up as:
If $10$%$\times x + x = y$
then $y - 10$%$\times y < x$

Is there a rate, used for both gain and depreciation, for which those two answers would actually be the same?
For the above reason, there is no such rate as this is true for all numbers and all percentages, except 0% and £0.

For a non-zero initial value, we cannot find a non-zero rate for which the time taken for the value to double and half would be the same. This is because for any rate, the ratio of values between two successive years will be different for gain and depreciation. To get the same time we really need these ratios to be the same.

Well done everyone!