1. N.Chandramowliswaran

2.  Let a,b,c be given positive integers.
Then ( abc, (a+b+c).(a2 + b2 + c2 ) ) = 1
----->(i)
iff ( a , b+c ) = 1
( b , c+a ) = 1
( c, a+b ) = 1
( a , b2 +c2 ) = 1
( b, c2 +a2 ) = 1
( c, a2 +b2 ) = 1

3. (i) is equivalent to
( abc, a+b+c ) = 1
and ( abc , a2 + b2 + c2 ) = 1
Hence the theorem

4.  Let a,b,c,d be four given positive integers.
Then ( (d.(a+c)) , ab + ac + ad + bc + bd + cd ) = 1 -->
iff ( a+c , bd+ac ) =1 --->(i)
( d, (a+c).b+ac) = 1 ---> (ii)

5.  Since is true,
There exists integers λ,μ
λ[ d.(a+c)]+μ[ab+ac+ad+bc+bd+cd] = 1
λ (a+c).d + μad + μbd + μcd + μab + μac + μbc = 1
(λ(a+c) + μa + μb + μc).d + μ[(a+c).b + ac] = 1
⇒(ii) is true

6. ⇒ (i) is True
Proof: λ ,μ are integers.
λ [ d. (a+c)] + μ[ab+ac+ad+bc+bd+cd] = 1
⇒ λd(a+c) + μ(b+d)(a+c)+ μ(bd+ac) = 1
⇒ [λd + μ (b+d)].(a+c) + μ(bd + ac) = 1

7. (a+c , bd+ac ) = 1 ---> (1)
(d, (a+c).b+ac) = 1---> (2)
(1)and (2) are true
Try to prove
(d. (a+c) , ab+ ac + ad +bc + bd+ cd ) = 1
Assume “p” is a prime
p| d.(a+c) and
p| ab +ac + ad+ bc+ bd+ cd
p| d. (a+c) ⇒ (p|d) V (p| a+c)
Suppose p|d
∴p| ab + ac+ bc
∴p| (a+c).b +ac
This is a contradiction for (d , (a+c).b+ac) = 1

8.  Suppose p| a+c
p| ab + ac + ad + bc + bd + cd
(i.e) p|(a+c)(b+d) + ac + bd
(i.e) p| ac + bd
Again this is a contradiction,
Since ( a+c , ac+bd ) = 1
Hence the theorem

9.  Suppose f: { P1 , P2 ,P3 ,P4 , P5 , P6 }
V(G)
Here “pi “ are distinct “very large” ODD primes
f (vi ) = Pi ∈ V(G)
1≤i≤6
( pj , ∑ pi pj ) = 1
Question: Can we find such a map f for this spanning tree “T” of “G”
ONTO
1-1
6
j=1 { vi ,vj } ∈E(T)