## Adding a digit in front of each number in: a^n + b^n + c^n = x^n + y^n + z^n, n=1,2

$9 \; + \; 5 \; + \; 4 \; = \; 8 \; + \; 7 \; + \; 3 \; = \; 18$
$9^2 \; + \; 5^2 \; + \; 4^2 \; = \; 8^2 \; + \; 7^2 \; + \; 3^2 \; = \; 122$

adding a digit in front of each number so that

$(10 \, a_1 + 9) + (10 \, b_1 + 5) + (10 \, c_1 + 4) \; = \; (10 \, x_1 + 8) + (10 \, y_1 + 7) + (10 \, z_1 + 3)$
$(10 \, a_1 + 9)^2 + (10 \, b_1 + 5)^2 + (10 \, c_1 + 4)^2 \; = \; (10 \, x_1 + 8)^2 + (10 \, y_1 + 7)^2 + (10 \, z_1 + 3)^2$

if we continue this process of adding a digit

$(10^2 \, a_2 + a_1 \, 9) + (10^2 \, b_2 + b_1 \, 5) + (10^2 \, c_2 + c_1 \, 4)$
$= \; (10^2 \, x_2 + x_1 \, 8) + (10^2 \, y_2 + y_1 \, 7) + (10^2 \, z_2 + z_1 \, 3)$

$(10^2 \, a_2 + a_1 \, 9)^2 + (10^2 \, b_2 + b_1 \, 5)^2 + (10^2 \, c_2 + c_1 \, 4)^2$
$= \; (10^2 \, x_2 + x_1 \, 8)^2 + (10^2 \, y_2 + y_1 \, 7)^2 + (10^2 \, z_2 + z_1 \, 3)^2$

and so on.

For example,

See if you can add more digits to the examples I posted.

Find other examples.

math grad - Interest: Number theory
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### 3 Responses to Adding a digit in front of each number in: a^n + b^n + c^n = x^n + y^n + z^n, n=1,2

1. Paul says:

There are many possibilities for this, for just single digit sums and squares alone there are 305 sets, from
13, 17, 18, 14, 15, 19

93, 97, 98, 94, 95, 99

then taking the first from that set we get 255 more with 3 digits, and again taking the first from that set we get 189 then 189 then 189 then 189 then189 etc (seems to now be constant at 189 solutions) I stopped at a 13 digit list with

1111111111113, 1665554433217, 1859637452318, 1231231212114, 1425314231215, 1979757553319
to
9111111111113, 9665554433217, 9859637452318, 9231231212114, 9425314231215, 9979757553319

So I’m guestimating there are 4523327798334374921458292775 possible 13 digit solution, this will of course contain all the reversals and permutations of digits.

Paul.

• Paul says:

I used the 100th solution from each set and worked up to 20 digits and the pattern was almost identical except the second set of solutions had 257 not 255, but all the others were the same, this is the last 20 digit solution where each preceding one was taken at 100.

95555555555555555443, 97777777777777777627, 93333333333333333598 and 97777777777777777514, 93333333333333333485, 95555555555555555669.

Paul.

• benvitalis says:

True, There are many possibilities. Thanks for posting these. In my next two blogs, I’ll post examples with palindromes and prime numbers.