TITLE: Is there a way to find a pythagorean triple so that when you place a given digit before it, it still is a pythagorean triple?
QUESTION [9 upvotes]: For example, in base 10:
$$5^2 + 12^2 = 13^2$$
And when I put a one before each number, the equality still holds:
$$15^2 + 112^2 = 113^2$$
So my question is, in a given base, is there a way to get pythagorean triples that still hold when a given digit precedes each?
Bonus question: And if there is, are there infinitely many?

REPLY [2 votes]: Inspired by Robert's analysis I decided to find a different family.
This works for bases of the form $b=4mn$ with $n>4$ and $m>0$.
Take the triple $(2mn,m(n^2-1),m(n^2+1))$
Clearly $2mn<4mn$ so its one digit long. Also $4mn<m(n^2-1)<m(n^2+1)<(16m^2n^2)$ so those two are two digits long.
If you put the digit $4m-1$ (which is a digit as $4m-1<4mn$) in front of each number you get the triple:
$$\bigg((4m-1)\cdot(4mn+2mn),(4m-1)\cdot(4mn)^2+m(n^2-1),(4m-1)\cdot(4mn)^2+m(n^2+1)\bigg)$$
$$=\bigg(16m^2n-2mn,64m^3n^2-16m^2n^2+mn^2-m,64m^3n^2-16mn^2+mn^2+m\bigg)$$
Which satisifies the Pythagorean Identity.
With $m=1$ this leads to Robert's solutions of:
$$\matrix{b = 20 & x = 10 & y = 24 & z = 26 & x' = 70 & y' = 1224 & z' = 1226\cr
b = 24 & x = 35 & y = 12 & z = 37 & x' = 1763 & y' = 84 & z' = 1765\cr
b = 28 & x = 14 & y = 48 & z = 50 & x' = 98 & y' = 2400 & z' = 2402\cr
b = 32 & x = 63 & y = 16 & z = 65 & x' = 3135 & y' = 112 & z' = 3137\cr
b = 36 & x = 18 & y = 80 & z = 82 & x' = 126 & y' = 3968 & z' = 3970\cr
b = 40 & x = 99 & y = 20 & z = 101 & x' = 4899 & y' = 140 & z' = 4901\cr}$$
And with $m=2$ this leads to solutions (Robert had first one) of:
$$\matrix{b = 40 & x = 20 & y = 48 & z = 52 & x' = 300 & y' = 11248 & z' = 11252\cr
b = 48 & x = 24 & y = 70 & z = 74 & x' = 360 & y' = 16198 & z' = 16202\cr
b = 56 & x = 28 & y = 96 & z = 100 & x' = 420 & y' = 22048 & z' = 22052\cr
b = 64 & x = 32 & y = 126 & z = 130 & x' = 480 & y' = 28798 & z' = 28802\cr
b = 72 & x = 36 & y = 160 & z = 164 & x' = 540 & y' = 36448 & z' = 36452\cr
b = 80 & x = 40 & y = 198 & z = 202 & x' = 600 & y' = 44998 & z' = 45002\cr}$$