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.

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TopNewesta 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.

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Nice

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@Aareyan Manzoor can you give the primitive solution!

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

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

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

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