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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
10 months, 1 week 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. Aareyan Manzoor · 10 months, 1 week ago

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@Aareyan Manzoor Nice Abdur Rehman Zahid · 9 months, 3 weeks ago

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

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@Sualeh Asif (2,3,1,2,2) @Sualeh Asif Aareyan Manzoor · 9 months, 3 weeks ago

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@Sualeh Asif i tried but couldnt find one. let me write a code, i will have a pair in a while... Aareyan Manzoor · 9 months, 3 weeks ago

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

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