Most of you know these triples:
$3: 4 :5$
$5: 12 :13$
$8: 15 :17$
$7: 24 :25$
$9: 40 :41$
More generally we can construct such triangles such as $$2x:x^2-1:x^2+1$$
My question is why one of the sides seems to be always prime? (When there is no common divisor)
$\endgroup$3 Answers
$\begingroup$It's great that you are interested and looking for patterns, so don't get discouraged. As one of my teachers is fond of saying,
If all your conjectures are true, you aren't trying hard enough.
So, here are first ten counterexamples:
$$\begin{array}{c|c} x& 2x:x^2-1:x^2+1\strut\\\hline 8 & 16:63: 65\strut\\ 12 & 24:143: 145\strut\\ 18& 36:323: 325\strut\\ 22& 44: 483: 485\strut\\ 28& 56: 783: 785\strut\\ 30& 60: 899: 901\strut\\ 32& 64: 1023: 1025\strut\\ 34& 68: 1155: 1157\strut\\ 38& 76: 1443: 1445\strut\\ 42& 84: 1763: 1765 \end{array}$$
Mathematica code:
listofcounterexamples = {};
For[x = 1, x < 100, x++, If[Length[listofcounterexamples] == 10, Break[]]; If[GCD[2x, x^2-1, x^2+1] == 1 && Not[PrimeQ[x^2+1]], AppendTo[listofcounterexamples, {x, 2x, x^2-1, x^2+1}]]];
listofcounterexamples $\endgroup$ $\begingroup$ The pattern you see (one side always having prime length) fails for larger $x$ and is a good example of the strong law of small numbers. Consider, for example, $x=12$.
$\endgroup$ $\begingroup$I'll add a comment on why your claim is probably false, even before you find any counterexamples. Your formula $$ 2x : x^2 - 1 : x^2 + 1 $$
should be a red flag. $2x$ is never prime for $x > 1$ because it is divisible by 2, and $x^2 - 1 = (x + 1)(x - 1)$ is never prime for $x > 2$ because it factors. So your claim seems dubious from the start -- only $x^2 + 1$ even has a chance of being prime.
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