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# Olympiad Corner

5) If $$a, b, c, d, e$$ are natural numbers, then prove that $$a^2 + b^3 + c^4 + d^5 = e^6$$ has infinite many solutions.

Note by Dev Sharma
2 years, 3 months ago

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## Comments

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a solution is (2,3,1,2,2). ssince there exists a solution, we multiply both sides by $$2^{60}$$ which gives us $(2^{30}a)^2+(2^{20}b)^3+(2^{15}c)^4+(2^{12}d)^5=(2^{10}e)^6$ we can repeat this process infinitly giving us infinite solutions. hence solved.

- 2 years, 3 months ago

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Nice

- 2 years, 3 months ago

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@Aareyan Manzoor can you give the primitive solution!

- 2 years, 3 months ago

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(2,3,1,2,2) @Sualeh Asif

- 2 years, 3 months ago

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i tried but couldnt find one. let me write a code, i will have a pair in a while...

- 2 years, 3 months ago

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$22^2+1^3+1^4+3^5=3^6$ Hence (22,1,1,3,3) is another solution

- 2 years, 2 months ago

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