Determine all possible list(s) of pythagorean triple (a,b,c), with 0 < a < b < c < 100, such that we will obtain another pythagorean triple (p,q,r) by inserting the same nonzero digit to the left of each of a, b and c. None of a, b and c can contain any leading zeroes.

Note: Try to derive a non computer assisted method, although computer program/spreadsheet solutions are welcome.

Since a,b, and c are all at most 2-digit numbers, there are only three cases to consider:

1) a,b,c have the same number of digits (1 or 2)
2) a and b have 1 digit but c has 2
3) a has 1 digit and b and c have 2

Case 1)

prepending the digit k gives the equation:
(a + 100k)^2 + (b + 100k)^2 = (c + 100k)^2 (for when a,b,c are 2-digit. Same logic holds for 1-digit except use 10k instead of 100k)

Expand these terms and subtract a^2+b^2 on the left and c^2 on the right (since a,b,c is a pythagorean triplet)

200ka + 200kb + 2*100*100*k^2 = 200kc + 100*100*k^2

Solve for k (which is not zero); divide everything by 200k and rearrange terms:

c + 50k = a + b + 100k
50k = c - (a+b)

but wait -- a,b,c are the sides of a triangle and so the sum of any two sides is greater than the third. Specifically, a+b>c and so
the right hand side is negative and k must be negative. We can't have a negative digit, so there are no solutions of the form in case 1)


Case 2)

Consider the numbers AFTER the addition of the digit k. The transformed a and b are each at most 100 and so the sum of their squares is at most 10,000. But c is at least 110 (being two digits prior to prefixing k) and so the transformed right side is always greater than the transformed left side. There are no solutions for case 2).

Case 3)

Using the same strategy as case 1), we now have:

(a+10k)^2 + (b + 100k)^2 = (c + 100k)^2

20ka + 100k^2 + 200kb + 100*100*k^2 = 200kc + 100*100*k^2

cancel identical terms and divide by 20k to get:

a + 5k + 10b = 10c
5k = 10(c-b) - a

k = 2(c-b) - a/5

Now, we've assumed that a is a single-digit number, and a>0, so for k to be an integer there's only one possibility: a=5. Only
one pythagorean triangle has shortest side 5 and that's (5,12,13). In this case, c-b=1, and k= 2(13-12) - 5/5 = 1.

(5,12,13) which becomes (15,112,113) is the only solution when a,b,c are all < 100.


As a generalization of this problem, I conjecture that ALL solutions to problems that involve
a pythagorean triplet (with no restrictions on number of digits) transforming into a new triplet by prefixing the same digit k to
all three members are of the form (5*10^j, 12*10^j, 13*10^j)

Posted by Paul on 2008-11-08 23:38:36