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

This is an interesting question with amazing concepts!
Hope you are well
Playlist to watch all videos on Learncommunolizer
• Maths Olympiad
And
• Maths
Subscribe: / @learncommunolizer

[@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.

[@MaratSafikhanov]

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

[@Nikioko]

Nope.

[@Евгений-ю2р]

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

[@meshulamoren8633]

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

[@AndresIsmael-ux7mp]

Al ojo a=5,b=4

[@user-cb7gy1xz3b]

a=5 , b= 4

[@THyperon]

5³-4³=61

[@stevenconnell3206]

5^3 - 4^3

[@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?

[@MaratSafikhanov]

🤗

[@shanmughansubramani7837]

64 equal to 125-64

[@shanmughansubramani7837]

Read 61

[@dardoburgos3179]

a= 5 y b= 4.

[@CharlesChen-el4ot]

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

[@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.