- Set
*f*(*x*) = 2*x*^{6}-6*x*^{4}-6*x*^{3}+12*x*^{2}+ 1 = 0 and*g*(*x*) = 2*x*^{6}-6*x*^{4}-4√2*x*^{3}+12*x*^{2}. By raising to the sixth power, we see that a solution to the given equation also satisfies*f*. Furthermore to have a real solution, we need*x*≤√2. Therefore if we can show that*f*(*x*) has no solutions with*x*≤√2, then it will follow that the original equation has no solutions. Now*g*(*x*) = 2*x*^{2}(*x*- √2)^{2}(*x*^{2}+2√2*x*+ 3). Thus*g*has zeros at 0 and √2 (of multiplicity 2), and is positive otherwise, because*x*^{2}+2√2*x*+ 3 > 0 for all*x*∈ℝ. Now*f*(*x*) -*g*(*x*) = (4√2 -6)*x*^{3}+ 1 which is positive for*x*≤√2, because the function is decreasing and (4√2 -6)√2^{3}+1 > 0. To see this, we need to show that 17 - 12√2 > 0. However multiplying by 17 + 12√2, we see that we need to show 17^{2}-144·2 > 0, which is true. It follows that the given equation has no real solutions. - Write
*t*= tan(*x*/2). Then cos^{2}(*x*/2) = 1/(1 +*t*^{2}), socosand since tan*x*= cos^{2}(*x*/2) - sin^{2}(*x*/2) = (1 -*t*^{2})/(1 +*t*^{2})*x*= 2*t*/(1 -*t*^{2}),sinWrite*x*= cos*x*tan*x*= 2*t*/(1 +*t*^{2}).*I*= ∫_{0}^{a}*dx*/(1 + cos*x*+ sin*x*). Since*dt*/*dx*= (sec^{2}(*x*/2))/2 = (1 +*t*^{2})/2, we see that*I*= ∫_{0}^{tan(a/2)}2*dt*/((1 +*t*^{2}) + (1 -*t*^{2}) + 2*t*) = ∫_{0}^{tan(a/2)}*dt*/(1 +*t*).*I*= ln(1 + tan(*a*/2)). (An alternative answer is ½ln(1 + sin*a*)/(1 + cos*a*) + ½ln 2.) When*a*= π/2, we have tan(*a*/2) = 1 and we deduce that*I*= ln 2 as required. - We may assume that
*AB*= 1. Since ∠*APB*= 150, the sine rule yields, sin 150/*AB*= sin 20/*AP*= sin 10/*BP*and sin 30/*AP*= sin 40/*CP*. Therefore*PC*= 4 sin 20 sin 40 = 2 cos 20 - 1. Write ∠*PBC*= θ. Since ∠*BPC*= 100, we see that ∠*PCB*= 80 - θ, and then the sine rule for triangle*BPC*yields(2 cos 20 - 1)/(sinθ) = 2 sin 10/(sin(80 - θ)) = (2 sin 10)/(cos(θ + 10)).Therefore2 cos 20 cos(θ +10) = 2 sin 10 sinθ + cos(θ +10) = cos(θ - 10).We deduce that cos(30 + θ) + cos(10 - θ) = cos(θ - 10) and hence cos(30 + θ) = 0. We conclude that θ = 60. - Denote the vertices of the triangle by
*A*,*B*and*C*(counterclockwise). Let*P*be an interior point of the triangle and draw lines parallel to the three sides, partitioning the triangle into three triangles and three parallelograms. Let*EH*be the segment parallel to*AC*, let*FI*be the segment parallel to*BC*, and let*JG*be the segment parallel*AB*. Here the points*E*,*F*lie on the edge*AB*; the points*G*,*H*lie on the edge BC, and the points*I*,*J*lie on the edge*AC*. Suppose that the area of the triangle*EFP*is*a*, the area of the triangle*PGH*is*b*, and the area of the triangle*JPI*is*c*. Note that the triangles*EFP*,*PGH*,*JPI*and*ABC*are similar. Therefore*EF*/*PG*= √*a*/√*b*and*JP*/*PG*= √*c*/√*b*. Thus (*EF*+*JP*)/*PG*= (√*a*+ √*c*)/√*b*and hence 1 + (*EF*+*JP*)/*PG*= 1 + (√*a*+ √*c*)/√*b*, i.e.(Since*PG*+*EF*+*JP*)/*PG*= (√*a*+ √*b*+ √*c*)/√*b*.*PG*=*FB*and*JP*=*AE*, because*FBGP*and*AEJP*are parallelograms,*AB*/*PG*= (√*a*+ √*b*+ √*c*)/√*b*. Because*ABC*is similar to*PGH*, we have*AB*/*PG*= √*T*/√*b*. Therefore √*T*= √*a*+ √*b*+ √*c*. - Let
(
*a*,*b*)∈*S*and let*d*= gcd(*a*,*b*). Then*a*=*dm*and*b*=*dn*with gcd(*m*,*n*) = 1. Since*g*(*a*,*b*)∈ℕ, we see that*ab*=*d*^{2}*mn*is a perfect square and hence*mn*is a perfect square. Therefore*m*and*n*are both perfect squares, because gcd(*m*,*n*) = 1. Thus we may write*a*=*ds*^{2}and*b*=*dt*^{2}with gcd(*s*,*t*) = 1.By assumption,

*h*(*a*,*b*) = 2*ds*^{2}*t*^{2}/(*s*^{2}+*t*^{2})∈ℕ. Since gcd(*s*^{2}+*t*^{2},*s*^{2}) = gcd(*s*^{2}+*t*^{2},*t*^{2}) = gcd(*s*^{2},*t*^{2}) = 1, it follows that*s*^{2}+*t*^{2}divides 2*d*. Thus*a*=*k*(*s*^{2}+*t*^{2})*s*^{2}/2 and*b*=*k*(*s*^{2}+*t*^{2})*t*^{2}/2 for some*k*∈ℕ.Now

*a*≠*b*because*s*≠±1. Also*f*(*a*,*b*) =*k*(*s*^{2}+*t*^{2})^{2}/4∈ℕ. We have two cases to consider.- If
*s*^{2}+*t*^{2}is odd, then 4|*k*and hence*f*(*a*,*b*)≥4(1^{2}+2^{2})^{2}/4 = 25. - If
*s*^{2}+*t*^{2}is even, then*s*and*t*are odd because gcd(*s*,*t*) = 1 and hence*f*(*a*,*b*)≥(1^{2}+3^{2})/4 = 25.

*f*(*a*,*b*)≥25. However*f*(5, 45) =*f*(10, 40) = 25, so the minimum of*f*over*S*is 25. - If
- Set
*g*(*x*) =*f*(*x*) -*x*^{2}+ 4*x*- 2. Then*g*(1) =*g*(4) =*g*(8) = 0. Therefore we may write*g*(*x*) = (*x*- 1)(*x*- 4)(*x*- 8)*q*(*x*) where*q*(*x*)∈ℤ[*x*]. Since*f*(*n*) =*n*^{2}- 4*n*- 18, we see that*g*(*n*) = -20 and hence (*n*- 1)(*n*- 4)(*n*- 8)*q*(*n*) = -20. By inspection,*n*= 3 or 6. We note that both of these values of*n*can be obtained, by taking (for example)*q*(*x*) = -2 and 1 respectively, and then*f*(*x*) = -2(*x*- 1)(*x*- 4)(*x*- 8) +*x*^{2}- 4*x*+ 2 and (*x*- 1)(*x*- 4)(*x*- 8) +*x*^{2}- 4*x*+ 2 respectively. - First we look at small values of
*n*: the given equation is a quadratic in*m*. If*n*∈{0, 1, 2, 4}, there are no solutions. If*n*= 3, then*m*= 6 or 9. If*n*= 5, then*m*= 9 or 54. We now proceed by contradiction to show that there is no solution if*n*≥6. So suppose (*m*,*n*) is a solution with*n*≥6. Then*m*divides 2·3^{n}and so either*m*= 3^{a}for some 0≤*a*≤*n*, or*m*= 2·3^{b}for some 0≤*b*≤*n*. If*m*= 3^{a}, then2On the other hand if^{n+1}-1 =*m*+ 2·3^{n}/3^{a}= 3^{a}+2·3^{n-a}.*m*= 2·3^{b}, then2Therefore there must be nonnegative integers^{n+1}-1 =*m*+ 2·3^{n}/*m*= 2·3^{b}+3^{n-b}.*a*,*b*such that2Note that 3^{n+1}-1 = 3^{a}+2·3^{b},*a*+*b*=*n*.^{a}< 2^{n+1}< 3^{2(n+1)/3}and 2·3^{b}< 2^{n+1}< 2·3^{2(n+1)/3}, because 3^{2/3}> 2. Thus*a*,*b*< 2(*n*+ 1)/3. Since*a*+*b*=*n*, we deduce that(Now let*n*- 2)/3 <*a*< 2(*n*+ 1)/3 and (*n*- 2)/3 <*b*< 2(*n*+ 1)/3.*t*= min(*a*,*b*). Then*t*> (*n*- 2)/3 and since*n*≥6, it follows that*t*> 1. Because 3^{t}divides 3^{a}and 2·3^{b}, we see that 3^{t}divides 2^{n+1}- 1. Since*t*≥2, we deduce that 2^{n+1}≡1 mod 9. Now 2^{n+1}≡1 mod 9 if and only if 6 divides*n*+ 1, so*n*+ 1 = 6*r*for some*r*∈ℕ. Therefore2Since 3^{n+1}-1 = 4^{3r}-1 = (4^{2r}+4^{r}+1)(4^{r}-1) = (4^{2r}+4^{r}+1)(2^{r}-1)(2^{r}+ 1).^{t}divides 2^{n+1}- 1, we see that 3^{t}divides (4^{2r}+4^{r}+1)(2^{r}-1)(2^{r}+ 1). Note that 9 does not divide 4^{2r}+4^{r}+ 1, hence 3^{t-1}divides (2^{r}-1)(2^{r}+ 1). Since gcd(2^{r}-1, 2^{r}+ 1) = 1, either 3^{t-1}| 2^{r}- 1 or 2^{r}+ 1. In any case, 3^{t-1}≤2^{r}+ 1. Then 3^{t-1}≤2^{r}+1≤3^{r}= 3^{(n+1)/6}. Therefore (*n*- 2)/3 - 1 <*t*- 1≤(*n*+ 1)/6. This yields*n*< 11, which is a contradiction, because*n*≥6 and we proved that 6 |*n*+ 1.

Peter Linnell 2017-10-28