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
·
1 year, 6 months ago

## Comments

Sort by:

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 · 1 year, 6 months ago

Log in to reply

– Abdur Rehman Zahid · 1 year, 5 months ago

NiceLog in to reply

@Aareyan Manzoor can you give the primitive solution! – Sualeh Asif · 1 year, 5 months ago

Log in to reply

@Sualeh Asif – Aareyan Manzoor · 1 year, 5 months ago

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

– Aareyan Manzoor · 1 year, 5 months 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 · 1 year, 5 months ago

Log in to reply