A Nice Math Olympiad Algebra Problem You Should Know

[@hanswust6972]

Very nice problem!

[@devondevon4366]

21 Assuming all the variables are positive integers, not equal to each other, then 'r' cannot be greater than 1 (since 2^2 * 16 =64 >48 ); hence r=1. Hence 16 r^2 = 16 Hence p^2 + 4 q^2 = 32; hence q is < than 3 (since 4* 3^2 =36 > 32); hence q= 2. Hence 4q^2 = 4*4=16; p^2 = 16 (32 -16); hence p=4. Hence p^2 + q^2 + r^2 = 16 + 4 + 1 = 21 Answer checking pq + 4qr + 2rp =24 4*2 + 4*2*1 + 2*1*4 = 8 + 8 + 8 =24

[@nasrullahhusnan2289]

Your assumption that p, q, and r not equal to each other is not necessary. That they are integers is sufficient assumption, and the second given equation is not needed. The answer will be (p,q,r)={(4,2,1),(-4,-2,-1)}

[@nasrullahhusnan2289]

Given: p²+4q²+16r²=48 p, q, and r are integers Take modulo 4: [mod(p,4)]²=0 mod(p,4)=0 It means that p=4k. Plugging to the given equation, after dividing by 4 we get 4k²+q²+4r²=12. Take modulo 4: [mod(q,4)]²=0 Hence q=4w where w=any integer. Plug it to the given equation and divide by 16: k²+(2w)²+r²=3. It means that k²=(2w)²=r²=1 p=4k --> p²=16 q²=(4w)² --> q²=4 r²=1 Hence p²+q²+r²=16+4+1=21 The second given equation.can be replaced by assumption that p, q, and r are integers.

[@abdoali6268]

Keep going ❤

[@manojitmaity7893]

If sinA + sin²A + sin³A = 1 ,then what is the value of cos ⁶A - 4cos⁴A +8cos ²A ?

[@DARK-mo7jw]

4

[@hans7831]

It's always 42

[@DARK-mo7jw]

@@hans7831 i think something wrong with your method or calculation. Please check it again. It's 4

[@manojitmaity7893]

@@DARK-mo7jw How ???

[@DARK-mo7jw]

Change all the values of cos in term of sin and then proceed

[@markobasej8636]

4