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1/a+1/b=1/13->(b+a)/(a*b)=1/13->a*b=13m & b+a=m, con m número entero positivo, de la última ecuación b=m-a, sustituyendo en la penúltima nos lleva a la ecuación a(m-a)=13m->am-a^2=13m->a^2-ma+13m=0, aplicando la "chicharronera" a=(m+-sqrt(m^2-4*13m))/2, para que a sea entero el discriminante m^2-52m debe ser un cuadrado perfecto, presentándose dos casos: Caso I m^2-52m=0, de donde m=0 o m=52, descartando m=0 nos queda m=52, obteniendose a=52/2=26 & b=52-26=26, de donde a+b=52. Caso II m^2-52m=m*(m-52)=m(m-4*13)=m*4*(m/4-13) es un cuadrado perfecto si m es un cuadrado perfecto múltilpo de 4 tal que m/4-13 sea a su vez cuadrado perfecto, como los cuadrados perfectos múltiplos de 4 son 4, 16, 36, 64, 100, 144, 196,...tomando m=196 obtenemos a=196/2+-2*sqrt(196)*sqrt(196/4-13)/2=98+-84, a=14 o a=182, pero en cualquiera de los casos a+b=196...¿Habrá otro múltiplo de 4 que cumpla con las condiciones?
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The answer is 52. And I better use that for refreshing my understanding!!!
good
The case II is unnecesary by symmetry, thank you
ok thanks
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There's no need to make all that mess. 13 in a prime number, so it can derive only from the sum of 1/26 + 1/26.
ok
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You said it short and sweet.. Good thinking👍
Amazing
thanks
1/a+1/b=1/13->(b+a)/(a*b)=1/13->a*b=13m & b+a=m, con m número entero positivo, de la última ecuación b=m-a, sustituyendo en la penúltima nos lleva a la ecuación a(m-a)=13m->am-a^2=13m->a^2-ma+13m=0, aplicando la "chicharronera" a=(m+-sqrt(m^2-4*13m))/2, para que a sea entero el discriminante m^2-52m debe ser un cuadrado perfecto, presentándose dos casos: Caso I m^2-52m=0, de donde m=0 o m=52, descartando m=0 nos queda m=52, obteniendose a=52/2=26 & b=52-26=26, de donde a+b=52. Caso II m^2-52m=m*(m-52)=m(m-4*13)=m*4*(m/4-13) es un cuadrado perfecto si m es un cuadrado perfecto múltilpo de 4 tal que m/4-13 sea a su vez cuadrado perfecto, como los cuadrados perfectos múltiplos de 4 son 4, 16, 36, 64, 100, 144, 196,...tomando m=196 obtenemos a=196/2+-2*sqrt(196)*sqrt(196/4-13)/2=98+-84, a=14 o a=182, pero en cualquiera de los casos a+b=196...¿Habrá otro múltiplo de 4 que cumpla con las condiciones?
Great job, thanks
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so ez bruh
yes, easy
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