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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
1 year, 4 months ago

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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. · 1 year, 4 months ago

Nice · 1 year, 3 months ago

@Aareyan Manzoor can you give the primitive solution! · 1 year, 3 months ago

(2,3,1,2,2) @Sualeh Asif · 1 year, 3 months ago

i tried but couldnt find one. let me write a code, i will have a pair in a while... · 1 year, 3 months ago

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