MO Problems

For Q1:
There's a simpler sol'n:
Consider (a - 1)^2 >= 0
we then have a^2 + 1 >= 2a ---(1)
similarly, b^2 + 1 >= 2b ---(2)

(1) + (2), i.e. a^2 + b^2 + 2 >= 2a + 2b
(a + b) +2 >= 2(a + b)

Thus, 2 >= a + b, Max.(a+b) = 2



PS. there's also a interesting sol'n by using linear programming.
the eq'n is actually a circle: a^2 + b^2 - a - b = 0
We have
Max f(a,b) = a+b
Sub. to a^2 + b^2 - a - b = 0

By sketching the graph, we can easily get a+b=2 is the optimal sol'n.

Find integers solution only , not find all integers solution , so no need to
consider the case when y is odd...

2013-02-27 21:32:23 補充：
It's better to consider the case when y is odd...

2013-02-27 21:34:49 補充：
Also it's better to consider the case when y < 0.

for question (1), have to consider the case when y is odd...

Please help with the followings
(1) If a^2+b^2=a+b，find themaximum of a+b
Sol
a^2+b^2=a+b
a^2－a+b^2－b=0
a=[1+/-√(1+4b－4b^2)]/2
2a=1+/-√(1+4b－4b^2)
2a+2b－1=2b+/-√(1+4b－4b^2)
When max a>=0，b>=0
2a+2b－1=2b+√(1+4b－4b^2)
f(b)=2b+√(1+4b－4b^2)，b>=0
f’(b)=2+(1/2)*[1/√(1+4b－4b^2)]*(4－8b)
Set f’(b)=0
2+(1/2)*[1/√(1+4b－4b^2)]*(4－8b)=0
1+(1－2b)*[1/√(1+4b－4b^2)]=0
1=(2b－1)*[1/√(1+4b－4b^2)]
√(1+4b－4b^2)]=2b－1
1+4b－4b^2=4b^2－4b+1
8b^2－8b=0
b=0 or b=1
(1) b=0
2a+0－1=0+√1
a=1
a+b=1
(2) b=1
2a+2－1=2+√(1+4－4)=3
a=1
a+b=2
綜合(1)，(2)
Maxa+b=2

(2) if x^2 +399 = 2^y， find integers solution for x andy.
Sol
p=｜x｜
y=2q
2^(2q)－p^2=399
(2^q－p)(2^q+p)=399
399=3*7*19 有8個正因數
1，3，7，19，21，57，133，399
(1) 2^q－p=1，2^q+p=399
2^(q+1)=400 (不合)
(2) 2^q－p=3，2^q+p=133
2^(q+1)=136 (不合)
(3) 2^q－p=7，2^q+p=57
2^(q+1)=64
q=5
p=25
x=+/-25，y=10
(4) 2^q－p=19，2^q+p=21
2^(q+1)=40 (不合)
綜合(１)，(2)，(3)，(4)
x=+/-25，y=10