Math Olympiad | A Nice Algebra Problem | a=? & b=?

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[@N_Kumar26]

As a^3 -b^3=61
a>b (Here difference of odd minus even is odd,61 is odd number)
a=2n+1
b=2n
n is positive integer
(2n+1)^3-(2n)^3 =61
8n^3+12n^2+6n+1-8n^3-61=0
2n^2+n-10=0
n=-5/2,2
a=2×2+1=5
b=2x2=4

[@sankariraman6453]

Easily solve by trial and error md125-64=61

[@MaratSafikhanov]

(5)³-(4)³=61

[@keescanalfp5143]

yeah and
(-4)³ - (-5)³ = +61 ?
and then reject these -4 and -5 , because, though possible, they are negative and not natural .

[@steakdemamouth2807]

An other solution: a =4 and b=square 3

[@keescanalfp5143]

could we just wish that
³√3 were still a natural number .

[@kennethkan3252]

a^3-b^3=61
a=2mod3 >4
a=+-(2,5,8,11,14)
b=1mod3 >1
b=+-(1,4,7,10,13)
a^3-b^3=61
5^3-4^3=61
(-5)^3-(-4)^3=61
a=5 b=4
a=-5,b=-4

[@-wx-78-]

5:24 Sum of roots should be -1, definitely.

[@meshulamoren8633]

a³=61+b³
a³=61+1=62
a³=61+2³=69
a³=61+3³=88
a³=61+4³=125
a=5
b=4

[@MaratSafikhanov]

(a,b)=(-4,-5; 5,4)

[@Nikioko]

Nope.

[@CharlesChen-el4ot]

😂a-b =1; 3ab=61-1=60
ab = 20
a = 5, b = 4

[@GeorgetaBobar]

a=5 , b= 4

[@Евгений-ю2р]

a^3-b^3+125-125-61=0 далее a^3-5^3=b^3-4^3 Ответ: a=5; b=4

[@MaratSafikhanov]

🤗

[@AndresIsmael-ux7mp]

Al ojo a=5,b=4

[@shanmughansubramani7837]

64 equal to 125-64

[@shanmughansubramani7837]

Read 61

[@kennethkan3252]

a^3-b^3=61
(-4)^3-(-5)^3=61
a=-4, b=-5

[@Nikioko]

Wrong.

[@kennethkan3252]

@@Nikioko
(-64)-(-125)=61
125-64=61

[@Nikioko]

@@kennethkan3252 Yes, but your answer is still wrong. Do you know what a definition set is? Do you know what natural numbers are?

[@dardoburgos3179]

a= 5 y b= 4.

[@THyperon]

5³-4³=61

[@stevenconnell3206]

5^3 - 4^3

[@MaratSafikhanov]

(-4)³-(-5)³=61

[@dieterjagutis6226]

-4 - 5 is Z set bu not in N set

[@Nikioko]

-4 and -5 are no natural numbers.