N.Chandramowliswaran

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

(i) is equivalent to

( abc, a+b+c ) = 1

and ( abc , a2 + b2 + c2 ) = 1

Hence the theorem

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)

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

⇒ (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

(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

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

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)