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.
–
Aareyan Manzoor
·
10 months, 1 week ago

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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. – Aareyan Manzoor · 10 months, 1 week ago

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– Abdur Rehman Zahid · 9 months, 3 weeks ago

NiceLog in to reply

@Aareyan Manzoor can you give the primitive solution! – Sualeh Asif · 9 months, 3 weeks ago

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@Sualeh Asif – Aareyan Manzoor · 9 months, 3 weeks ago

(2,3,1,2,2)Log in to reply

– Aareyan Manzoor · 9 months, 3 weeks ago

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

\[22^2+1^3+1^4+3^5=3^6\] Hence (22,1,1,3,3) is another solution – Abdur Rehman Zahid · 9 months, 1 week ago

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