Vol. 7 No 5 - Pi Mu Epsilon

Proof.
2 2
Suppose p~ q (mod 3); then 3lp-q and 3lp -q . But
p-qSpt2q (mod 31, so 3 \p+2q. We also have 3 l(p-q)+(p+2q)=2p+q, and hence
2 2 2
3 1 2 ~ ~ Therefore t ~ . (p -q , 2pq+q )#l.
2 2 2
Now suppose (p -q , 2pqtq ) = a' $1. Let x be any prime divisor
of d. Now x divides neither p nor q; this follows since xip
2
-q
2
and
hence if x divides one it also divides the other, contradicting (p,q)=l.
2 2 2
Since xbpq+q we must have x\ 2p+q. Also x\p -q implies x\p-q or xlptq.
If xl2ptq and xlp+q, then xlp, a contradiction.
Hence x12ptq and xlp-q
and these in turn imply x 13p. Thus x=3 and 3 1 p -q making p E q (mod 3).
To complete the proof of Theorem 1 note that by Lemma 3 it is
enough to show that any two of k,l,m are relatively prime.
gives us (k,l)=l.
Lemma 6 now
Next we prove the converse to Theorem 1. Define (k,Z,m) to be
a trivial solution triple if it is of the form (-k,k,k) or (k,O,k).
7hn.oit.e.m 2. If (k,l,m) is a non-trivial primitive solution triple,
then there exist unique integers p and q with p>0, (p,q)=l and pfq (mod 3)
such that k=p
2
-q
2
, ~ .=2~~tq, and m=p
2
tpq+q
2
.
since (k.2.m)
2 2 2 2 2
Proof. Since k +kM =m we have Z(ktZ)=m -k =(m+k)(m-k), and
is non-trivial we can write
2 m-k
= mt = = t
where t is a non-zero rational number.
equations:
2-kt=& , Ztt(t+l)k=m .
Solving this system for k and 2 we get
This gives us a system of
k=--, (1-t2)m
2 = (t2+2t)m
t2tttl
t 2 tt+l
Now since t is rational, we let t = q/p where p and q are integers with
p>0 and (p,q)=l. Clearly the p and q satisfying these conditions are
unique.
Substituting for t we get
Lemma 5 implies that 2-kf.m (mod 3) and thus Zfmtk (mod 3). These
facts together with (p,q)=l imply p&7 (mod 3). Lemmas 3 and 6 now tell
2 2 2
us that (p2-q2, p2+pqtq2)=l and (2pq+q2, 2 p2tpqtq 2 )=I. Thus p +pqtq . lm,
and since (k,Z)=l we must in fact have p +pq+q =m.
This concludes the proof of Theorem 2.
Summarizing the previous results we have
Thus we can wr'it
2 2 2
The.ote.rn 3. All integer triples (k,Z,m) for which k tkZ+k =m
and k-Zzm (mod 3) are either:
(1) trivial, that is of the form (-k,k,k) or (k,O,k)
(2) non-trivial, that is, there exist unique integers p,q,v
2 2
withp>0, q#O, v#O, (p,q)=l, and p fq (mod 3) such that k=(p -q )r ,
2 2 2
2=(2pq+q 1, and m=(p +pq+q )r.
2 2 2
To find all solutions to k -kZtl =m we can use Theorem 3 and
Lemmas 1 and 2.
First note that primitivity is defined as before and
that any solution triple is a multiple of a primitive one.
The next
lemma indicates that there are essentially two distinct families of
primitive solutions.
Lennna 7.
If the triple of integers is a primitive solution to
k2-k2+12=m2, then k+Z%m (mod 3) or kt2 -m (mod 31, but not both.
The proof is similar to the proof of Lemma 5 and is omitted.
2 2 2
Observe that unlike the k +k2+2 =m problem we cannot choose
which congruence we want to hold, since interchanging k and 2 leaves
both unaffected.
are either
(O,k,k),
Combining Lemma 2 with Theorem 3 we have:
2 2 2
Theolem 4. All integer triples (k,l,m) for which k -kl+l =m
(1) trivial, that is of the form (-k.0.k). (k,O.k), (k,k,k),
(4, -k,k), or (0,-k,^;
(2) non-trivial, that is, there exist unique integers p,q,Y
with p>0, q#O, &O,
(p,q)=l, and ptq (mod 3) such that
k = (p2-q2)m 2 = ( 2pqtq2 )m
p2+ Pa +q2 p2+ to2 '
Recall that Z/(m+k) = t = q/p, and by assumption k-2s m (mod 3).
It is an easy matter to see that the first non-trivial family
corresponds to ktZs-m(mod 3) while the second corresponds to k+23 (mod 3).