Solve for b to get b = (44-a^2) / (1+2a). The denominator is positive for all non-negative a. Since b is non-negative, it means that the numerator must be non-negative. Hence, we only need to test the values a=0,1,2,3,4,5, and 6 and see which ones give integer values for b.
a^2 has to be < 44, therefore a can only be 1,2,3,4,5 or 6. For a=1: 44-1= 3*b no solution in Z. For a=2: 44-2^2=5*b b=8 For a=3: 44-3^2=7*b. b=5 For a=4: 44-4^2=9*b no solution in Z For a=5: 44-5^2=11*b. no solution in Z For a=6: 44-6^2=13*b. no solution in Z. There are two pairs that solve the equation: a=2; b=8 and a=3; b=5
You made it the hard way a and b are positive integer a² + 2ab + b = 44 (a+b)² - a² - 2ab - b² + a² + 2ab + b = 44 (a+b)² + b - b² = 44 (a +b)² > 44 If (a +b)² = 49, b - b² = -4, roots are not integers If (a +b)² = 64, b - b² = -20, b ≠ -4, b = 5, hence a = 3 If (a +b)² = 81, b - b² = -37, roots are not integers If (a +b)² = 100, b - b² = -56, b ≠ -7, b = 8, hence a = 2
I thought of b=8 a=2 2ab= 32 a^2= 4 . This adds 4+32+8= 44. But I started from: suppose (a+b)(a+b)= 10x10=100 You can start from somewhere of your choosing and come up with answers of all sorts. Did not see any restrictions on validity of answers.
If a, b є Z+, then (once eliminating the 2a on left and right) you are left with 1 < 4b-1, which considering that the minimum value of b in Z+ is 1, is true for all b.
b = (44-a²)/(2a+1) and (2a+1)>0. If (2a+1) > (2a+4b-1), then (in this case) a > 6 which causes b will negative. But the condition is that a and b are positive. So (2a+1) must be less than (2a+4b-1)
Solve for b to get b = (44-a^2) / (1+2a).
The denominator is positive for all non-negative a.
Since b is non-negative, it means that the numerator must be non-negative.
Hence, we only need to test the values a=0,1,2,3,4,5, and 6 and see which ones give integer values for b.
a^2 has to be < 44, therefore a can only be 1,2,3,4,5 or 6.
For a=1: 44-1= 3*b no solution in Z.
For a=2: 44-2^2=5*b b=8
For a=3: 44-3^2=7*b. b=5
For a=4: 44-4^2=9*b no solution in Z
For a=5: 44-5^2=11*b. no solution in Z
For a=6: 44-6^2=13*b. no solution in Z.
There are two pairs that solve the equation: a=2; b=8
and a=3; b=5
If a=0, b=44.
@ the solution has to be in Z+(as mentioned in the very beginning), therefore Zero is no solution.
@@fuerdichundmich You're right. I was thinking that it was positive, but it is neutral.
You made it the hard way
a and b are positive integer
a² + 2ab + b = 44
(a+b)² - a² - 2ab - b² + a² + 2ab + b = 44
(a+b)² + b - b² = 44
(a +b)² > 44
If (a +b)² = 49, b - b² = -4, roots are not integers
If (a +b)² = 64, b - b² = -20, b ≠ -4, b = 5, hence a = 3
If (a +b)² = 81, b - b² = -37, roots are not integers
If (a +b)² = 100, b - b² = -56, b ≠ -7, b = 8, hence a = 2
I thought of b=8 a=2 2ab= 32 a^2= 4 . This adds 4+32+8= 44.
But I started from: suppose (a+b)(a+b)= 10x10=100
You can start from somewhere of your choosing and come up with answers of all sorts. Did not see any restrictions on validity of answers.
Yo, why did we asume that 2a+1
If a, b є Z+, then (once eliminating the 2a on left and right) you are left with 1 < 4b-1, which considering that the minimum value of b in Z+ is 1, is true for all b.
b = (44-a²)/(2a+1) and (2a+1)>0.
If (2a+1) > (2a+4b-1), then (in this case) a > 6 which causes b will negative. But the condition is that a and b are positive. So (2a+1) must be less than (2a+4b-1)