If a, b, c and d are natural number such that a5= b6=c3=d4 and d-a = 61 then smallest value of c-b
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Let a5=b6=x30a5=b6=x30 and c3=d4=y12c3=d4=y12.
Then
d−a=y3−x6⇒(y−x2)(y2+yx2+x4)=61d−a=y3−x6⇒(y−x2)(y2+yx2+x4)=61.
Since 61 is prime, y−x2=1y−x2=1 and y2+yx2+x4=61y2+yx2+x4=61.
Solving, y=5,x=2y=5,x=2.
Then c−b=593c−b=593.
Now, why would I choose to solve mine instead of yours. Because yours forces x=yx=y which is not true. Conclusion, yours has no solution as stated
Then
d−a=y3−x6⇒(y−x2)(y2+yx2+x4)=61d−a=y3−x6⇒(y−x2)(y2+yx2+x4)=61.
Since 61 is prime, y−x2=1y−x2=1 and y2+yx2+x4=61y2+yx2+x4=61.
Solving, y=5,x=2y=5,x=2.
Then c−b=593c−b=593.
Now, why would I choose to solve mine instead of yours. Because yours forces x=yx=y which is not true. Conclusion, yours has no solution as stated
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