Number theory ultimatum by Rohith, Anirudh and Arjun!
a
2
+
b
2
+
c
2
=
2
(
a
+
b
+
c
+
1
)
and
a
b
+
b
c
+
c
a
=
a
2
+
b
2
+
c
2
−
c
Let a , b , c are positive integers which satisfy the equations above, find a + b + c .
The answer is 6.
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2 solutions
main bhi yehi kiya... it's simple method if provided they (a,b and c) are integers!
Put the value of a 2 + b 2 + c 2 from equation 1 in eq. 2 we get;- a b + c ( a + b ) = 2 ( a + b ) + c + 2 this furher can be written as; a b + ( c − 2 ) ( a + b − 1 ) = 4 ; a b < = 4 from equation 1 we see that two of a,b,c are odd and one even. Or all even so for ab = 4 , a = 2 , b =2 . and this doesn't satisfy the first equation. Similar manipulation for ab = 1 and 3 gives no solution . For ab = 2 , a = 1 and b = 2 and this also gives c = 3. This is a consistent solution as it satisfies both the equations therefore ( a + b + c ) = 6
could you please try to give a more detailed manner for solving this question. Although your solution is not wrong it doesn't seem to be encompassing.
Equation 1 can be factorised into (a-1)^2 + (b-1)^2 +(c-1)^2= 5. From here it is trivial that a,b,c are 1,2,3